Question 6
Triangle FGH and two of its angle measures are shown below.
G
31°
600
F
H
What is the measure of angle F?
A
29°
B
89°
с
90°
D 91°

Answer:

37.5%

Step-by-step explanation:

Answer:

Penny is taller than Jeremy.

Step-by-step explanation:

1 inch=2.54 cm

? = height of Jeremy

Make proportion:

1    - 2.54

38 -  ?

?= 38*2.54 ÷ 1

?= 96.52

Jeremy is 96.52 inches

96.52<100

Answer:

b,c,e,d,a,f

Step-by-step explanation:

Answer: m∠T = 48°

ΔMAT has MA = MT

=> ΔMAT is a isosceles triangle at M

=> ∠A = ∠T = (180° - ∠M)/2

=> x + 20  = (180 - 3x)/2

⇔ 2x + 40 = 180 - 3x

⇔ 5x = 140

⇔ x = 140/5

⇔ x = 28

=> ∠A = ∠T = 28 + 20 = 48°

Step-by-step explanation:

Answer: Correct answer is Side

Step-by-step explanation:

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

The point of intersection is found by equating the expressions for P:

  6d = 5d +15

  d = 15

  p = 6d = 6·15 = 90

The point of intersection of the graphs is (15, 90).

The first equation graphs as a line with slope 6 through the origin.

The second equation graphs as a line with slope 5 through point (0, 15).

It is inconvenient to graph p and d using the same scale, so the attache graph has the vertical scale compressed somewhat.

The correct expression to represent the similarity of these two triangles is,

21/35 = x/30.

A triangle is a three-sided closed-plane figure formed by joining three noncolinear points. Based on the side property triangles are of three types they are Equilateral triangle, Scalene triangle, and Isosceles triangle.

We know two triangles are similar if the ratio of their corresponding sides is equal.

Therefore, The correct way to represent the similarity is,

21/35 = x/30.

We can solve for x here,

35x = 21×30.

5x = 3×30.

x = 3×6.

x = 18.

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Answer:

x2√x

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Answer:

Step-by-step explanation:

First sequence : 10,9.5,9,8.5,...

Let's see the difference between 2 numbers :
10 - 9.5 = 0.5
9.5 - 9 = 0.5
9 - 8.5 - 0.5

So on and so on. So, the common difference between these numbers are 0.5

Next sequence : 200,100,50,25, ...

We see, to get from one number to another, we divide it by 2 / multiply it by 1/2.

200 x 1/2 = 100
100 x 1/2 = 50
50 x 1/2 = 25

So on and so on. So, the common ratio of these numbers are 1/2.

Answer:

Step-by-step explanation:

A common difference is the number you need to get from a chosen number on your left to get to the next number going right.

In the first sequence, you start with 10 on your left.

To get to the next number going right, you need to subtract 0.5 or add - 0.5

In any event, the common difference is -0,5

The second problem is a little  harder. It does not follow the same sort of pattern that the first question does. This time you are multiplying or dividing.

200 * 1/2 = 100

100 * 1/2 = 50

50 *  1/2 = 25

25 *  1/2 = 13.5

So the common ratio is 1/2

Answer:

12x+5y

Step-by-step explanation:

Combine 3x and 9x to get 12x.

  hope it helped  :)

The value of x from the given expression (3^1/2 ) ( 3^1/2 ) = 3^6 can be expressed as 6.

A formula that expresses the connection between two expressions on each side of a sign. Typically, it has a single variable and an equal sign. Like this: 2x - 4 Equals 2. In the above example, the variable x exists.

Given that (3^1/2 ) ( 3^1/2 ) = 3^6

then 3^(x/2+ x/2) = 3 ^6

x/2+ x/2 = 6

Then the lcm , we have

x+x/2 = 6

2x/2 = 6

2x = 12

x = 12/2 = 6

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Greek number prefixes are employed in the naming of molecular (covalent) substances.

Greek number prefixes are employed in the naming of molecular (covalent) substances. Prefixes are always combined in order of size (left-to-right in the table below). For instance, "heptadecatetracta" (7 + 10 + 400) would be the prefix for the number 417.

We utilise prefixes (mono-, di-, tri-, etc.) at the beginning of each element in molecular compounds to denote the element's subscript, but we only use the prefix mono- for the compound's second element.

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Answer:

All you have to do is find the diameter and divide it by 2. I'm sorry if I can't give you a direct answer because I cannot see the diagram.

Step-by-step explanation:

Answer:

7=2.75

8=18

9=3.625

10=29

Step-by-step explanation:

Answer:

Fard

Step-by-step explanation:

Hehe funi

B. Richard is more likely to draw two cards with a product that is a single digit because there are more single-digit numbers than even numbers.

To determine which outcome is more likely, we need to analyze the number of possible outcomes for each case.

For two cards to have a product that is an even number, we need at least one of the cards to be an even number. In a standard deck of playing cards, there are 26 even-numbered cards (2, 4, 6, 8, 10) and 22 odd-numbered cards (Ace, 3, 5, 7, 9).

Therefore, the number of possible outcomes for two cards with an even product is 26 multiplied by 52 (the total number of cards in two stacks) minus the 26 cases where both cards are odd.

On the other hand, for two cards to have a product that is a single digit, the possibilities are limited to the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9. In a deck of playing cards, only the Ace, 2, 3, 4, and 5 cards satisfy this condition.

Therefore, the number of possible outcomes for two cards with a single-digit product is 5 multiplied by 52 minus the 5 cases where both cards are not within the single-digit range.

By comparing the number of possible outcomes, we can see that the number of outcomes for two cards with a single-digit product is greater than the number of outcomes for two cards with an even product. Thus, Richard is more likely to draw two cards with a product that is a single digit.

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(C): Richard is more likely to draw two cards with a product that is an even number, because...

To determine the probabilities, let's consider the possible outcomes for each case:

Two cards with a product that is an even number:

To get an even product, at least one of the cards drawn must be even. There are three even cards in each stack (2, 4, and 6) and one odd card (3). The possible combinations are:

Even card × Even card

Even card × Odd card

Odd card × Even card

The probability of drawing an even card from the first stack is 3/4, and from the second stack is also 3/4. Since these events are independent, we multiply the probabilities:

P(Even card) = (3/4) × (3/4) = 9/16

Two cards with a product that is a single digit:

To get a product that is a single digit, both cards drawn must be either 1 or 2. There are two 1s and two 2s in each stack. The possible combinations are:

1 × 1

1 × 2

2 × 1

2 × 2

The probability of drawing a 1 from the first stack is 2/4, and from the second stack is also 2/4. Again, since these events are independent, we multiply the probabilities:

P(Single digit) = (2/4) × (2/4) = 4/16

Comparing the probabilities:

P(Even card) = 9/16

P(Single digit) = 4/16

Since 9/16 is greater than 4/16, Richard is more likely to draw two cards with a product that is an even number. Therefore, the correct answer is C: Richard is more likely to draw two cards with a product that is an even number.

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Answer:

x=4.4

Step-by-step explanation:

9x+2 = 4x+ 24

5x+2 = 24

5x = 22

5x/5 = 22/5

x=4.4

i got one solution scream at me if im wrong

Answer:

CE = 15

CD = 10

Step-by-step explanation:

DE is 1/3 of the total line

DE = 5

CE is 3/3 of the line 5 × 3 = 15 which is your whole line

CD is 2/3 of the line so you would need to 5 × 2 = 10

\(-5x = 78 + 25 <=> -5x = 103 <=> x = -103/5\)

The paths of two particles are described by parametric equations. The first particle follows a circular path, while the second particle follows a path with both circular and linear components.

We need to graph the paths and determine the number of points of intersection.

Additionally, we need to find the collision point where the particles occupy the same position at the same time.

Lastly, we need to determine if there is a collision when the x-coordinate of the second particle is modified.

(a) To graph the paths of both particles, we plot the parametric equations x1 = 3sin(t), y1 = 2cos(t) and x2 = -3 + cos(t), y2 = 1 + sin(t) on a coordinate plane for 0 ≤ t ≤ 2π. The paths of the particles will be represented by curves. By analyzing the graph, we can count the number of points of intersection.

(b) To find the collision point, we need to find the values of t where x1 = x2 and y1 = y2 simultaneously. By setting 3sin(t) = -3 + cos(t) and 2cos(t) = 1 + sin(t), we can solve for t. The obtained value(s) of t will give us the collision point (x, y) where the particles occupy the same position at the same time.

(c) If the x-coordinate of the second particle is modified to x2 = 3 + cos(t), we need to repeat the process of finding the collision point. By setting 3sin(t) = 3 + cos(t) and 2cos(t) = 1 + sin(t), we solve for t. Depending on the solution(s) of t, we can determine if there is still a collision or not.

Please note that since this question involves graphing and solving equations, it is best to draw the graphs and solve the equations visually or using numerical methods to obtain specific values.

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Answer:

--------------

First, find the sum of all the integers in the given range and then divide it by the total number of integers.

The sum of an arithmetic sequence can be found using the formula:

To find n, we use the formula:

Now, we can find the sum:

Finally, we find the average by dividing the sum by the number of integers:

Answer:

Area = 50 cm²

Step-by-step explanation:

Area of a trapezium ABDE = ½(a + b)h

where,

a = BD = 2.5 cm

b = AE = 7.5 cm

h = DE = ?

We need to find DE ti be able to calculate the area.

DE = CE - CD

DE = CE - 5

Since ∆ACE is similar to ∆BCD, therefore:

AE/BD = CE/CD

Plug in the values

7.5/2.5 = CE/5

Cross multiply

CE*2.5 = 7.5*5

CE*2.5 = 37.5

Divide both side any 2.5

CE = 37.5/2.5

CE = 15

Thus, DE = CE - 5 = 15 - 5

DE = 10

✅Area of trapezium = ½(BD + AE)DE

= ½(2.5 + 7.5)*10

= ½(10)10

= 50 cm²

PLS GIVE ME BRAINLIEST

hope this helps and have a good day

Answer:

y = 25(1.2)^t

Step-by-step explanation:

Initial population = 25

The general exponential function is expressed as;

y = y0(1+r)^t

y0 is the initial population

at t = 1, r = 20% = 0.2

y = 25(1+0.2)^t

y = 25(1.2)^t

Hence the required equation is y = 25(1.2)^t

The root of the equation \(\(f(x) = x^3 - 3x - 1\)\) between -1 and 1, after the second iteration of the False Position method, is approximately -1.

The False Position method involves finding the x-value that corresponds to the x-intercept of the line passing through \(\((a, f(a))\)\) and \(\((b, f(b))\),\)where (a) and (b) are the endpoints of the interval.

Let's begin the iterations:

Iteration 1:

\(\(a = -1\), \(f(a) = (-1)^3 - 3(-1) - 1 = -3\)\)

\(\(b = 1\), \(f(b) = (1)^3 - 3(1) - 1 = -3\)\)

The line passing through (-1, -3) and (1, -3) is (y = -3). The x-intercept of this line is at (x = 0).

Therefore, the new interval becomes [0, 1] since the sign of f(x) changes between\(\(x = -1\) and \(x = 0\).\)

Iteration 2:

\(\(a = 0\), \(f(a) = (0)^3 - 3(0) - 1 = -1\)\)

\(\(b = 1\), \(f(b) = (1)^3 - 3(1) - 1 = -2\)\)

The line passing through\(\((0, -1)\) and \((1, -2)\) is \(y = -x - 1\)\). The x-intercept of this line is at (x = -1).

After the second iteration, the new interval becomes [-1, 1] since the sign of f(x) changes between (x = 0) and (x = -1).

Therefore, the root of the equation \(\(f(x) = x^3 - 3x - 1\)\) between -1 and 1, after the second iteration of the False Position method, is approximately -1.

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A probability mass function is a function that describes the probability of a discrete random variable taking on each possible value in its sample space.

A probability mass function (PMF) is a function that assigns a probability to each possible value of a discrete random variable. The PMF maps each value to the probability of that value occurring in the random variable's sample space. The probabilities assigned by the PMF must sum to 1, since the random variable must take on one of the possible values in its sample space.

The PMF is used to calculate the expected value, variance, and other statistical properties of the random variable. It is also used to create probability distributions and generate simulations of the random variable. Probability mass functions are commonly used in probability theory, statistics, and other fields that involve discrete random variables.

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The dotplot above shows the percentage of air in 14 popular brands of chips. We are given a task to find the range of the distribution, calculate and interpret the standard deviation.

find the interquartile range, interpret this value, and calculate the range, standard deviation, and IQR for the other 13 bags of chips. Find the range of the distribution The range of the distribution is the difference between the maximum and minimum data values. To find the range, we first arrange the percent of air in ascending order.19, 28, 39, 41, 45, 46, 47, 48, 50, 50, 50, 59.

Range = 59 – 19

= 40. The range of the distribution of percent of air is 40.2.

Calculate and interpret the standard deviation The standard deviation is a measure of the amount of variation or dispersion of a set of values. It shows how much the values deviate from the mean of the distribution. We can calculate the standard deviation for the percent of air in chips using a calculator or software.

The formula for the standard deviation is:

\(σ = √ [ Σ ( xi – μ )2 / N ]\) where σ is the standard deviation, xi is each data value, μ is the mean of the distribution, and N is the number of data values.

Using a calculator, we get:

μ = 44.64,

σ = 11.13. Interpretation: The standard deviation of 11.13 shows that the percent of air in chips varies by an average of 11.13% from the mean value of 44.64%.3. Find the interquartile range. Interpret this value The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1) of the data set. We can find the IQR by first finding Q1 and Q3. To do this, we use a calculator or software that can calculate quartiles. Using a calculator, we get:

Q1 = 41,

Q3 = 50,

IQR = Q3 – Q1

= 9. Interpretation: The IQR of 9 shows that the middle 50% of the percent of air in chips ranges from 41% to 50%.4.

The dotplot suggests that the bag of Fritos chips, with only 19% of air, is a possible outlier. Recalculate the range, standard deviation, and IQR for the other 13 bags of chips. Compare these values with the ones you obtained in Questions 1 through 3.

Explain why each result makes sense. After excluding Fritos from the data set, we can recalculate the range, standard deviation, and IQR. The new data set contains 13 values. Here are the percent of air in chips in ascending order.28, 39, 41, 45, 46, 47, 48, 50, 50, 50, 59.1. The new range is 31.2. The new standard deviation is 8.93.

The new IQR is 9.Each value is smaller than the original value obtained in Questions 1 through 3. This is because Fritos has a very low percentage of air, which pulls the range, standard deviation, and IQR down. Excluding Fritos, the distribution of percent of air is less spread out and more clustered around the mean value.

To summarize, we found that the range of the distribution of percent of air is 40. The standard deviation of 11.13 shows that the percent of air in chips varies by an average of 11.13% from the mean value of 44.64%. The IQR of 9 shows that the middle 50% of the percent of air in chips ranges from 41% to 50%. Excluding Fritos, the range is 31, the standard deviation is 8.93, and the IQR is 9. The values are smaller than the original values because Fritos has a very low percentage of air.

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In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.