Step-by-step explanation:
1. 4^2 = 16
2. 32/16 = 2
3. 2*3 = 6
4. 17-6=11
5. 11 + 6 = 17
The answer is 17
WEEK 2 Direction: Answer the following problems. 1. Jun wanted to know how much ice cream he got in on scoop. The radius of a scoop is 2 inches. Find the volum Use 3.14 for pi. (SHOW YOUR SOLUTION) a) What is asked in the problem? b) What are the given facts? c) What is the formula to be used? d) Number Sentence e) Final answer. (show your solution pls)
We are given the radius of the scoop and asked to find the volume of ice cream in one scoop. By using the formula for the volume of a sphere and substituting the given radius, we can calculate the volume. The final answer is approximately 33.49 cubic inches.
a) The problem asks for the volume of ice cream in one scoop.
b) The given fact is that the radius of the scoop is 2 inches.
c) The formula to be used is the volume of a sphere, which is given by V = (4/3)πr³, where V is the volume and r is the radius.
d) Number Sentence:
- Given: Radius (r) = 2 inches
- Formula: V = (4/3)πr³
- Substituting the value: V = (4/3)π(2)³
- Simplifying: V = (4/3)π(8)
- Evaluating: V = (4/3)(3.14)(8)
- Multiplying: V = 33.49333333 (approx.)
e) Final answer: The volume of ice cream in one scoop is approximately 33.49 cubic inches.
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The 6th grade class at Elm City College Prep increased
by 5% this year. If there were 80 scholars in 6th grade
last year, how many are there this year? Show your wo
Answer:
84 students
Step-by-step explanation:
Let's suppose that 100% was last year's percentage of students.
100% = 80
1% = 80 ÷ 100 = 0.80
100 + 5 = 105% (This year)
0.80 x 105 = 84
There is another method to this called percentage increase, but I kind of forgot the formula. Hope this method is also correct for the question.
Please someone help, give the right answer it’s important
Answer:
(3) ∠RNQ ≅ ∠ LPM
Step-by-step explanation:
Based on the image uploaded;
line MP is congruent to line NQ
line ML is congruent to line RQ
line LP is congruent to line RN
From line MP which is congruent to line NQ, we can say that ∠RNQ ≅ ∠ LPM. That is, angle N of triangle RQN is congruent to angle P of triangle LMP
Thus, option 3 is the correct answer ( ∠RNQ ≅ ∠ LPM)
Express cos M as a fraction in simplest terms.
\(\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=\sqrt{c^2 - o^2} \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{30}\\ a=\stackrel{adjacent}{MN}\\ o=\stackrel{opposite}{18} \end{cases} \\\\\\ MN=\sqrt{ 30^2 - 18^2} \implies MN=\sqrt{ 576 }\implies MN=24 \\\\[-0.35em] ~\dotfill\\\\ \cos(M )=\cfrac{\stackrel{adjacent}{24}}{\underset{hypotenuse}{30}} \implies \cos(M)=\cfrac{4}{5}\)
A box of 11 transistors has 4 defective ones.
A) If 2 transistors are drawn from the box together, what is the probability that both transistors are defective?
B) If 2 transistors are drawn from the box together, what is the probability that neither transistor is defective?
C) If 2 transistors are drawn from the box together, what is the probability that one transistor is defective?
Answer:
(a) 0.1325
(b) 0.4045
(c) 0.463
Step-by-step explanation:
Let X denote the number of defective transistors.
The proportion of defective transistors is, p = 4/11 = 0.364.
All the transistors are independent of the others.
The random variable X follows a binomial distribution.
(a)
Compute the probability that both transistors are defective, if 2 transistors are drawn from the box together as follows:
\(P(X=2)={2\choose 2}(0.364)^{2}(1-0.364)^{2-2}\\\\=1\times 0.132496\times 1\\\\=0.132496\\\\\approx 0.1325\)
(b)
Compute the probability that neither transistors are defective, if 2 transistors are drawn from the box together as follows:
\(P(X=0)={2\choose 0}(0.364)^{0}(1-0.364)^{2-0}\\\\=1\times 1\times 0.404496\\\\=0.404496\\\\\approx 0.4045\)
(c)
Compute the probability that one transistors are defective, if 2 transistors are drawn from the box together as follows:
\(P(X=1)={2\choose 1}(0.364)^{1}(1-0.364)^{2-1}\\\\=2\times 0.364\times 0.636\\\\=0.463008\\\\\approx 0.463\)
What is the shape of the cross section of the figure that is perpendicular to the triangular bases and passes through a
vertex of the triangular bases?
A
a parallelogram that is not a rectangle
O a rectangle
O a triangle that must have the same dimensions as the bases
O a triangle that may not have the same dimensions as the bases
Answer:
a triangle that may not have the same dimensions as the bases
Step-by-step explanation:
The cross section of the figure that is perpendicular to the triangular bases and passes through a vertex of the triangular bases would be a triangle that may not have the same dimensions as the bases.
plz show work how u got the answer to be marked as brainlyest
Sorry this isnt an answer, but i need more information to be able to answer this question! there isnt a photo or any more text attached
Answer
the answer is 91 miles per hour
Step-by-step explanation:
divide 728 by 8
n this equation, one-fourth is added to the variable y.
y+14=34
What is the value of y?
The value of y is = -------------
Answer:
I think its 19.75
The base of the mountain is 6,500 feet above sea level and AB measures 230 feet across. Given that the measurements for QAP is 20° and QBP is 35°, how far above sea level is peak P ? Express your answer to the nearest foot.
Height above sea level:
Answer:
6610
Step-by-step explanation:
We have tan(X) = opposite/ adjacent
tan(QBP) = PQ/BQ
tan(35) = PQ/BQ ---eq(1)
tan(QAP) = PQ/AQ
tan(20) = \(\frac{PQ}{AB +BQ}\)
\(=\frac{1}{\frac{AB+BQ}{PQ} } \\\\=\frac{1}{\frac{AB}{PQ} +\frac{BQ}{PQ} } \\\\= \frac{1}{\frac{230}{PQ} + tan(35)} \;\;\;(from\;eq(1))\\\\= \frac{1}{\frac{230 + PQ tan(35)}{PQ} } \\\\= \frac{PQ}{230+PQ tan(35)}\)
230*tan(20) + PQ*tan(20)*tan(35) = PQ
⇒ 230 tan(20) = PQ - PQ*tan(20)*tan(35)
⇒ 230 tan(20) = PQ[1 - tan(20)*tan(35)]
\(PQ = \frac{230 tan(20)}{1 - tan(20)tan(35)}\)
\(= \frac{230*0.36}{1 - 0.36*0.7}\\\\= \frac{82.8}{1-0.25} \\\\=\frac{82.8}{0.75} \\\\= 110.4\)
PQ = 110.4
≈110
Height above sea level = 6500 + PQ
6500 + 110
= 6610
URGENT!! HELP PLEASE!!
Linear programming problems can also be solved using a graphing calculator. The following instructions will work for most commonly used graphing calculators, but note that they might vary slightly based on calculator model. Now that you’ve learned how to use a graphing calculator to solve linear programming problems, solve the following problems.
Part A
Consider the given system
{2x+y ≤20
{3x+2y ≤30
{x,y ≥0
Graph the inequalities on your graphing calculator, and find the vertex points of this system
Part B
In part A, you obtained the vertex points of the given system. Test these vertex points in the objective function, f(z)=4x+6y using a graphing calculator. Find the maximum point.
Answer:
A. see below for a graph
B. f(x, y) = f(0, 15) = 90 is the maximum point
Step-by-step explanation:
A. See below for a graph. The vertices are those defined by the second inequality, since it is completely enclosed by the first inequality: (0, 0), (0, 15), (10, 0)
__
B. For f(x, y) = 4x +6y, we have ...
f(0, 0) = 0
f(0, 15) = 6·15 = 90 . . . . . the maximum point
f(10, 0) = 4·10 = 40
_____
Comment on evaluating the objective function
I find it convenient to draw the line f(x, y) = 0 on the graph and then visually choose the vertex point that will put that line as far as possible from the origin. Here, the objective function is less steep than the feasible region boundary, so vertices toward the top of the graph will maximize the objective function.
The vertex points of this system are (0, 0), (0, 15) and (10, 0). And the maximum point of the system is 90.
What are systems of inequalities?At least two linear inequalities in the same variables make up a system of linear inequality in two variables. The graph of a linear inequality is the graph of all solutions to the system, and the solution of a linear inequality is the ordered pair that resolves all inequalities in the system.
The given system:
2x+y ≤20
3x+2y ≤30
x, y ≥0
To find the vertex points,
first, we take inequalities in slope-intercept form.
y = 20 - 2x
And y = -3x/2 + 15
x, y ≥0
From that we interpreted the graph.
And the graph is given in the attached image.
A:
Check out the graph below. Given that the first inequality entirely encloses it, the vertices are those defined by the second inequality: (0, 0), (0, 15), (10, 0).
B:
We have objective function f(z) = 4x + 6y
To find the maximum point:
Substituting (0,0) to the function,
4x + 6y
= 4(0) + 6(0)
= 0
Substituting (0,15) to the function,
4x + 6y
= 4(0) + 6(15)
= 0 + 90
= 90
Substituting (10,0) to the function,
4x + 6y
= 4(10) + 6(0)
= 40 + 0
The maximum point is 90.
Therefore, (0, 0), (0, 15) and (10, 0) are the vertex of the system. And the maximum point is 90.
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For an investment of 26,245, a quarterly statement reports that the account is 26,292. The statement reports that for the same quarter, the rate of return of the investment was - 0.02%. Given the information regarding the investment quarterly activity, does the reported rate of return seem reasonable?
No. The reported return rate appears lower than expected.
Rate of returnsTo determine whether the reported rate of return is reasonable or not, we can calculate the expected rate of return based on the investment amount and the ending value of the account.
First, we need to calculate the total number of quarters that have elapsed between the initial investment and the current quarter. Let's assume that the investment has been held for n quarters.
Then, we can use the following formula to calculate the expected rate of return:
Expected rate of return = (Ending value / Initial investment) ^ (1/n) - 1
Plugging in the given values, we get:
n = 1 (since we are only given information for a single quarter)
Initial investment = 26,245
Ending value = 26,292
Expected rate of return = (26,292 / 26,245) ^ (1/1) - 1 = 0.18%
Comparing the expected rate of return of 0.18% with the reported rate of return of -0.02%, we can see that they are quite different. However, it's important to note that a single quarter's rate of return may not be indicative of the overall performance of the investment.
Therefore, while the reported rate of return seems lower than expected, we would need to consider the performance over a longer period of time to make a more informed assessment of the investment's performance.
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You have 48 minutes to exercise at the gym. The elliptical trainer burns 9 calories per minute, while the treadmill burns 8 calories per minute. You want to burn 400 calories total using both machines. How much time should you spend on each machine
Answer:
16 minutes on the elliptical trainer and 32 minutes on the treadmill.
Step-by-step explanation:
We have the following relation for the training in both machines:
\( 400 cal = 9 cal/min*t_{1} + 8 cal/min*t_{2} \) (1)
Where:
t₁: is the time spent on the elliptical trainer
t₂: is the time spent on the treadmill
With:
\( t_{1} + t_{2} = 48 min \) (2)
From equation (2) we have:
\( t_{1} = 48 min - t_{2} \) (3)
By entering equation (3) into (1):
\( 400 cal = 9 cal/min*(48 min - t_{2}) + 8 cal/min*t_{2} \)
\( 400 cal = 432 cal - 9 cal/min*t_{2} + 8 cal/min*t_{2} \)
\( t_{2} = \frac{32 cal}{1 cal/min} = 32 min \)
Now, we can find t₁:
\( t_{1} = 48 min - t_{2} = 48 min - 32 min = 16 min \)
Therefore, you should spend 16 minutes on the elliptical trainer and 32 minutes on the treadmill.
I hope it helps you!
Find the domain and range of the following function ƒ(x) = 5|x - 2| + 4 Domain: [4,8) Range: (-∞,∞) Domain: (4,∞) Range: (-∞,∞) Domain: (-∞,∞) Range: [4,∞) Domain: (-∞,∞) Range: (4,∞)
Answer:
Step-by-step explanation:
Hi,
the function is defined for all reals so the domain is \(]-\infty;+\infty[\)
for x real
|x| >= 0
so f(x) >= 4
so the range is \([4;+\infty[\)
do not hesitate if you need any further explanation
hope this helps
Answer:
Domain: (-∞,∞) Range: (4,∞)
point
Daniel can mow lawns at a constant rate of 3/5 lawns/hour.
How many lawns can Daniel mow in 10 hours?
f1c96ca233d8e9261bd3d1e241 ad7c1d webm 158 KB
~
Ob
Daniel can mow 6 lawns in 10 hours.
Daniel can mow 3 lawns in 10 hours,
Daniel can mow 5 lawns in 10 hours.
Answer:
6 lawns
Step-by-step explanation:
In one hour, Daniel can mow 3/5 lawns
therefore, in 10hours Daniel can mow 3/5 x 10 lawns = 6 lawns
Topic: Rates and proportions
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In each of the following figures, pll. Find the indicated values.12(2x-3)P13513(2x-3)P10314XEP21 = (8x + 10)22 = (6x + 18)92122
Given data:
It is given that p and q are parallel.
Now, since p and q are parallel so we have,
\(\begin{gathered} 2x-3=135\text{ (corresponding angles)} \\ 2x=138 \\ x=69 \end{gathered}\)So,
\(x=69\)Given f(x) = x² - 5x - 6 and g(x) =
x² - 6x, what are the domain restrictions
for (-)(x)?
A) x # +6
B) x = 0
C) x = 0,6
D) x = -2,6
The domain restrictions for (f/g)(x) are (c) x = 0, 6
How to determine the domain restrictions for (f/g)(x)?From the question, we have the following parameters that can be used in our computation:
f(x) = x² - 5x - 6
g(x) = x² - 6x
The composite function (f/g)(x) is calculated as
(f/g)(x) = f(x)/g(x)
substitute the known values in the above equation, so, we have the following representation
(f/g)(x) = (x² - 5x - 6 )/(x² - 6x)
For the domain restriction, we have
x² - 6x = 0
When solved, we have
x = 6 or x = 0
Hence, the domain restrictions for (f/g)(x) are (c) x = 0, 6
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Question
Given f(x) = x² - 5x - 6 and g(x) = x² - 6x, what are the domain restrictions
for (f/g)(x)?
A) x = +6
B) x = 0
C) x = 0,6
D) x = -2,6
14. y = -(x - 2)² + 1 graphed
Two-variable linear equations have a line as their graph, which is why they are referred to as linear equations. Plotting (x1,y1) and (x2,y2), then creating a line that connects the two.
How do you draw a graph ?\($\frac{d}{d x}\left(-(x-2)^2+1\right)$\)
Domain of \($-(x-2)^2+1:\left[\begin{array}{cc}\text { Solution: } & -\infty < x < \infty \\ \text { Interval Notation: } & (-\infty, \infty)\end{array}\right]$\)
Range of \($-(x-2)^2+1:\left[\begin{array}{cc}\text { Solution: } & f(x) \leq 1 \\ \text { Interval Notation: } & (-\infty, 1]\end{array}\right]$\)
Interception zones on the axis \($-(x-2)^2+1: \quad X$\) Intercepts: \($(3,0),(1,0), Y$\)Intercepts: (0,-3) .
Vertex of \($-(x-2)^2+1: \quad$\) Maximum (2,1) .
The graph of the equation y=2x+1 shows a straight line, or it is a linear equation. When the value of x rises, eventually the value of y rises by twice the rise in x plus one. With a ruler and the specified amount as the starting point, draw a vertical line. From the line across to the -axis, create a horizontal line using a ruler. Look at the value on the -axis.
There are three fundamental ways to graph linear functions. The initial method involves charting points and then tracing a line through them. The second is by using the slope and y-intercept. The identity function f(x)=x is transformed as the third method.
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the equation a=0.003x^2+21.3 models the average ages of women when they first married since the year 1940. In this equation, a represents the average age and x represents the years since 1940. Estimate the year in which the average age of brides was the youngest
Answer:
Please help me important question in image
Step-by-step explanation:Please help me important question in image
Please help me important quePlease help me important question in image
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Please help me important question in image
Please help me important question in image
Answer:
The equation a=0.003x^2+21.3 models the average ages of women when they first married since the year 1940 in the United States. In this equation, a represents the average age and x represents the years since 1940. To estimate the year in which the average age of brides was the youngest, we need to find the minimum value of the quadratic function a=0.003x^2+21.3. This can be done by using the formula x=-b/2a, where b is the coefficient of x and a is the coefficient of x^2. In this case, b=0 and a=0.003, so x=-0/(2*0.003)=0. This means that the average age of brides was the lowest when x=0, which corresponds to the year 1940. The value of a when x=0 is a=0.003*0^2+21.3=21.3, so the average age of brides in 1940 was 21.3 years old. This is consistent with the historical data, which shows that the median age of women at their first wedding in 1940 was 21.5 years old. The average age of brides has been increasing since then, reaching 28.6 years old in 2021.
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A mechanic charges C(t) dollars to install a new distributor in a car. The formula C(t)=16t+160 is used to find the cost of installation, where t is the number of hours of labor performed by the mechanic. What does the slope represent?
Answer:
Labor cost per hourStep-by-step explanation:
The given formula is linear, with the slope of 16 and the y-intercept 160.
The slope represents a labor cost per hour and the y-intercept is the cost of spare parts used in repair.
A box has a width of 10 cm and a length of 17 cm. The volume of the box is decreasing at a rate of 527 cubic cm per minute, with the width and length being held constant. What is the rate of change, in cm per minute, of the height when the height is 6 cm?
Round your answer to the nearest hundredth. (Do not include any units in your answer.)
Therefore, the rate of change, in cm per minute, of the height when the height is 6 cm is approximately -6 cm/min.
Given,The width of the box = 10 cm Length of the box = 17 cmThe volume of the box = 527 cubic cm/minWe need to find the rate of change, in cm per minute, of the height when the height is 6 cm.We know that the volume of the box is given as:V = l × w × h where, l, w and h are length, width, and height of the box respectively.It is given that the width and length are being held constant.
Therefore, we can write the volume of the box as
:V = constant × h Differentiating both sides with respect to time t, we get:dV/dt = constant × dh/dtNow, it is given that the volume of the box is decreasing at a rate of 527 cubic cm per minute.
Therefore, dV/dt = -527.Substituting the given values in the above equation, we get:
527 = constant × dh/dt
We need to find dh/dt when h = 6 cm.To find constant, we can use the given values of length, width and height.Substituting these values in the formula for the volume of the box, we get:
V = l × w × hV = 17 × 10 × hV = 170h
We know that the volume of the box is given as:V = constant × hSubstituting the value of V and h, we get:
527 = constant × 6 cm
constant = 87.83 cm/minSubstituting the values of constant and h in the equation, we get
-527 = 87.83 × dh/dtdh/dt = -6.0029 ≈ -6 cm/min
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Fill in the blanks
Transfer of heat energy around the Earth from uneven heating of its surface is accomplished by _____ air _____ patterns.
Transfer of heat energy around the Earth from uneven heating of its surface is accomplished by global air circulation patterns.
Answer:
Global
Circulation
Step-by-step explanation:
Hope this helps! :)
plz help?
Find the value of x in this proportion.
621=10x
Answer:
x = 62.1
Step-by-step explanation:
Note the equal sign in this equation, what you do to one side, you do to the other. Isolate the variable, x, by dividing 10 from both sides of the equation:
\(621 = 10x\\\\\frac{621}{10} = \frac{10x}{10} \\\\x = \frac{621}{10} \\x = 62.1\)
Answer:
x = 62.1
Step-by-step explanation:
What is the measure of each interior angle of the regular polygon pictured below? If necessary, round to the nearest tenth.
\(\underset{in~degrees}{\textit{sum of all interior angles}}\\\\ n\theta = 180(n-2) ~~ \begin{cases} n=\stackrel{number~of}{sides}\\ \theta = \stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ n=5 \end{cases}\implies 5\theta =180(5-2) \\\\\\ 5\theta =180(3)\implies 5\theta =540\implies \theta =\cfrac{540}{5}\implies \theta =108\)
A bucket contains six white balls and five red balls. A sample of four balls is selected
at random from the bucket, without replacement. What is the probability that the
sample contains...
Exactly two white balls and two red balls?
At least two white balls?
To solve this problem, we can use the formula for probability:
P(event) = number of favorable outcomes / total number of outcomes
First, let's find the total number of outcomes. We are selecting 4 balls from 11 without replacement, so the total number of outcomes is:
11C4 = (11!)/(4!(11-4)!) = 330
where nCr is the number of combinations of n things taken r at a time.
Now let's find the number of favorable outcomes for each part of the problem.
Part 1: Exactly two white balls and two red balls
To find the number of favorable outcomes for this part, we need to select 2 white balls out of 6 and 2 red balls out of 5. The number of ways to do this is:
6C2 * 5C2 = (6!)/(2!(6-2)!) * (5!)/(2!(5-2)!) = 15 * 10 = 150
So the probability of selecting exactly two white balls and two red balls is:
P(2W2R) = 150/330 = 0.45 (rounded to two decimal places)
Part 2: At least two white balls
To find the number of favorable outcomes for this part, we need to consider two cases: selecting 2 white balls and 2 red balls, or selecting 3 white balls and 1 red ball.
The number of ways to select 2 white balls and 2 red balls is the same as the number of favorable outcomes for Part 1, which is 150.
To find the number of ways to select 3 white balls and 1 red ball, we need to select 3 white balls out of 6 and 1 red ball out of 5. The number of ways to do this is:
6C3 * 5C1 = (6!)/(3!(6-3)!) * (5!)/(1!(5-1)!) = 20 * 5 = 100
So the total number of favorable outcomes for selecting at least two white balls is:
150 + 100 = 250
And the probability of selecting at least two white balls is:
P(at least 2W) = 250/330 = 0.76 (rounded to two decimal places)
Craig likes to collect vinyl records. Last year he had 10 records in his collection. Now he has 12 records. What is the increase of his collection
Answer:
2
Step-by-step explanation:
Please help I’ll mark you as brainliest if correct!
Create an expression that has the same value as (6x-4) + (x + 5).
Write the correct numbers from the list in the blank boxes. Each number
may be used once, more than once, or not at all.
Answer: 7x + 1
Step-by-step explanation:
(6x-4)+(x+5)
Step 1: Remove Parentheses:
6x-4+x+5
Step 2: Combine Like Terms:
7x +1
Please solve and find the equation and put the equation in quadratic form
To find the solution to the question we would use the general representation of a quadratic equation below:
\(ax^2+bx+c=y\)We then pick a suitable point
When x=0 and y =16
\(\begin{gathered} a(0)^2+b(0)+c=16 \\ c=16 \end{gathered}\)We then pick two more suitable points, then solve the resulting simultaneous equation
When x= 3 and y= 0
we have
\(\begin{gathered} a(3)^2+b(3)+16=0 \\ 9a+3b=-16 \end{gathered}\)Also, when x=-5 and y =0
we have
\(\begin{gathered} a(-5)^2+b(-5)+16=0 \\ 25a-5b=-16 \\ \end{gathered}\)We have gotten two equations, we will solve them simultaneously using the elimination method.
We will multiply equation one by 5 and equation two by 3
\(\begin{gathered} 5(9a+3b=-16)_{} \\ 3(25a-5b=-16) \end{gathered}\)\(\begin{gathered} 45a+15b=-80 \\ 75a-15b=-48 \\ Add\text{ equation 2 from 1} \\ 120a=-128 \\ a=-\frac{128}{120} \\ a=\frac{-16}{15} \end{gathered}\)Substitute the value of a in equation 1
\(\begin{gathered} 45(-\frac{16}{15})+15b=-80 \\ 3\times(-16)+15b=-80 \\ -48+15b=-80 \\ 15b=-80+48 \\ 15b=-32 \\ b=-\frac{32}{15} \end{gathered}\)We will then substitute a and b and c in the representation of the quadratic equation.
\(\begin{gathered} -\frac{16}{15}x^2-\frac{32}{15}x+16=0 \\ Multiply\text{ through by 15} \\ -16x^2-32x+240=0 \\ therefore\text{ we have} \\ -x^2-2x+15=0 \end{gathered}\){(2,-4),(3,-3),(4.-2),(3,-1),(2,0)}
Chris says that this relation represents a function jazzmyn says that it is not a function who do u agree with
Answer:
jazzmyn
Step-by-step explanation:
it is not a function because there are repeat x values. 3 shows up in two different coordinates as the x value.
how many hundred billions in the number 432,845,979,484
There are 4 hundred billions in the number 432,845,979,484.
To determine how many hundred billions are in the number 432,845,979,484, we need to consider the place value of the hundred billions digit.
The number 432,845,979,484 has 12 digits, and we count the digits from right to left, starting with the units place. The digit in the hundred billions place is the seventh digit from the right.
Let's break down the number:
4 hundred billions (4 x 100,000,000,000)
3 ten billions (3 x 10,000,000,000)
2 billions (2 x 1,000,000,000)
8 hundred millions (8 x 100,000,000)
4 ten millions (4 x 10,000,000)
5 millions (5 x 1,000,000)
9 hundred thousands (9 x 100,000)
7 ten thousands (7 x 10,000)
9 thousands (9 x 1,000)
4 hundreds (4 x 100)
8 tens (8 x 10)
4 units (4 x 1)
As we can see, there are 4 hundred billions in the number 432,845,979,484.
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