Part A REASONING The length of a rectangle is twice the width. Find the width if the perimeter is 60 centimeters. a. Select a variable and an equation for finding the width of the rectangle. A) w = length; 2(2w) + w = 60 B) W = width: 2w + 2w = 60 C) w = width; 2(2w) + 2w= 60 O D) w = width; 2w + w = 60​

Answer:

Angle 2

Step-by-step explanation:

Find S and find X.

Then find U.

The angle at X connected to S and U is the angle number.

Answer:

x=45

Step-by-step explanation:

i hope its right good luck :)

i think im wrong

The error that Janice made is that she did not convert the improper fraction StartFraction 130 over 30 EndFraction to a mixed number correctly in her final answer.

The error that Janice made can be identified by analyzing her calculations. Let's examine each step she took:

Step 1: 10 and StartFraction 5 over 6 EndFraction divided by 1 and two-thirds

In this step, Janice correctly converted the mixed number to an improper fraction, which becomes 10 and StartFraction 5 over 6 EndFraction = StartFraction 65 over 6 EndFraction.

Step 2: StartFraction 65 over 6 EndFraction divided by five halves

In this step, Janice attempted to divide the fractions. However, she made an error by adding the numerators and denominators instead of multiplying. This is incorrect. The correct operation for dividing fractions is to multiply the first fraction by the reciprocal of the second fraction. The reciprocal of five halves is two fifths.

Step 3: StartFraction 65 over 6 EndFraction times StartFraction 2 over 5 EndFraction

Here, Janice multiplied the fractions correctly. The product of these fractions is StartFraction 130 over 30 EndFraction.

Step 4: StartFraction 130 over 30 EndFraction = 4 and one-third

In this step, Janice attempted to convert the improper fraction to a mixed number. However, she made another error in her calculation. The fraction StartFraction 130 over 30 EndFraction is not equal to 4 and one-third. The correct conversion would yield 4 and two-sixths or 4 and one-third.

Therefore, the error that Janice made is that she did not convert the improper fraction StartFraction 130 over 30 EndFraction to a mixed number correctly in her final answer.

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Answer:

the independent variable is the lawn size

Step-by-step explanation:

Answer: The lawn size

Step-by-step explanation: The independent variable is the input which causes the output to change.

Answer:

Diagonals are perpendicular

1 diagonal bisects the other

Opposite angles are congruent

i) 4 more gallons of petrol will be needed to complete the journey.

ii) The cost of the petrol for the 420 km journey is GH¢55.00.

i) To determine the number of gallons of petrol needed to complete the journey, we can calculate the total distance that can be covered with the available petrol and then subtract it from the total distance of the journey.

Given that the car consumes 1 gallon of petrol for every 30 km, we can calculate the distance that can be covered with 10 gallons of petrol by multiplying 10 (gallons) by 30 (km/gallon):

Distance covered with 10 gallons = 10 * 30 = 300 km

To find the remaining distance that needs to be covered, we subtract the distance covered with the available petrol from the total distance of the journey:

Remaining distance = Total distance - Distance covered with available petrol

Remaining distance = 420 km - 300 km = 120 km

Since the car consumes 1 gallon of petrol for every 30 km, we can determine the additional gallons of petrol needed by dividing the remaining distance by 30:

Additional gallons needed = Remaining distance / 30 = 120 km / 30 km/gallon = 4 gallons

Therefore, the driver will need 4 more gallons of petrol to complete the journey.

ii) To calculate the cost of the petrol for the journey of 420 km, we need to multiply the total number of gallons used for the journey by the cost per gallon.

Given that a gallon of petrol costs GH¢5.50, and the total number of gallons used for the journey is 10 (given in the problem), we can calculate the cost using the formula:

Cost of petrol = Total gallons used * Cost per gallon

Cost of petrol = 10 gallons * GH¢5.50/gallon = GH¢55.00

Therefore, the cost of the petrol for the journey of 420 km is GH¢55.00.

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Answer:

x = 65

Step-by-step explanation:

the sum of the interior angles of a quadrilateral = 360°

sum the angles and equate to 360

x + 95 + 118 + 82 = 360

x + 295 = 360 ( subtract 295 from both sides )

x = 65

The simplification of the expressions in the question using the rules of indices are as follows;

1.) 6·e⁰ + (11·f)⁰ - 5/g⁰ = 2

2. (3⁻⁴ + 5⁻³)⁻¹ = \(49\frac{31}{206}\)

3. (3⁻⁴ + 5⁻³)⁻² = \(2415\frac{32685}{42436}\)

4.) \(\dfrac{-5\cdot \left(m^{-4}\cdot n^{-5} \right)^{-3}}{10\cdot c^{-5} \cdot d^6 \cdot e^{-8}}\) = \(\dfrac{c^5\cdot e^8}{2\cdot d^5}\)

5.)  \(\dfrac{-5\cdot \left(m^{-4}\cdot n^{-5}\right)^{-3 }}{7\cdot \left(p^{-6} \cdot q^8\right)^{-4}} =\dfrac{-5\cdot m^{12}\cdot n^{15}\cdot q^{32}}{7\cdot p^{24}}\)

An index or indices is the power by which a number or variable or expression raised.

1.) 6·e⁰ + (11·f)⁰ - 5/g⁰

6·e⁰ + (11·f)⁰ - 5/g⁰ = 6 × 1 + 1 - 5/1

6 × 1 + 1 - 5/1 = 6 + 1 - 5 = 2

Therefore;

6·e⁰ + (11·f)⁰ - 5/g⁰ = 2

2.) (3⁻⁴ + 5⁻³)⁻¹

(3⁻⁴ + 5⁻³)⁻¹ = (1/(3⁴) + 1/(5³))⁻¹ = 1/(1/(3⁴) + 1/(5³))

1/(1/(3⁴) + 1/(5³)) = 1/(1/81 + 1/125) = 1/((125 + 81)/(81×125))

1/((125 + 81)/(81×125)) = (81 × 125)/(125+81) = 10125/209

10125/206 = 49 + 31/206

(3⁻⁴ + 5⁻³)⁻¹ = \(49 \frac{31}{206}\)

3.) (3⁻⁴ + 5⁻³)⁻²

(3⁻⁴ + 5⁻³)⁻² = 1/((1/3⁴ + 1/5³)²)

1/((1/3⁴ + 1/5³)²) = 1/((1/81 + 1/125)²)

1/((1/81 + 1/125)²) = 1/(((125 + 81)/(81 × 125))²)

1/(((125 + 81)/(81 × 125))²) = 1/((206/10125)²)

1/((206/10125)²) = (10125)²/(206)² = 102515625/42436

102515625/42436 = 2415 32685/42436

(3⁻⁴ + 5⁻³)⁻² = \(2415\frac{32685}{42436}\)

4.)  \(\dfrac{5\cdot (2\cdot a^{-1}\cdot b^3)^0}{10\cdot c^{-5}\cdot d^6\cdot e^{-8}}\)

\(\dfrac{5\cdot (2\cdot a^{-1}\cdot b^3)^0}{10\cdot c^{-5}\cdot d^6\cdot e^{-8}} = \dfrac{5\times 1}{10\times \left(\frac{1}{c^5}\times d^6 \times \frac{1}{e^8} \right) }\)

\(\dfrac{5\times 1}{10\times \left(\frac{1}{c^5}\times d^6 \times \frac{1}{e^8} \right) } = \dfrac{5}{10\times \left(\frac{d^6}{c^5\times e^8} \right) }\)

\(\dfrac{5}{10\times \left(\frac{d^6}{c^5\times e^8} \right) } = \dfrac{1}{2\times \left(\frac{d^6}{c^5\times e^8} \right) }\)

\(\dfrac{1}{2\times \left(\frac{d^6}{c^5\times e^8} \right) } = \dfrac{c^5\times e^8}{2\cdot d^6}\)

\(\dfrac{5\cdot (2\cdot a^{-1}\cdot b^3)^0}{10\cdot c^{-5}\cdot d^6\cdot e^{-8}}= \dfrac{c^5\times e^8}{2\cdot d^6}\)

5.) \(\dfrac{-5\cdot \left(m^{-4}\cdot n^{-5} \right)^{-3}}{7\cdot \left(p^{-6}\cdot q^8\rright)^{-4}}\)

\(\dfrac{-5\cdot \left(m^{-4}\cdot n^{-5} \right)^{-3}}{7\cdot \left(p^{-6}\cdot q^8\right)^{-4}} =\dfrac{-5\cdot \dfrac{1}{\left(\frac{1}{m^{4}}\times \frac{1}{n^{5}} \right)^{3}} }{7\cdot \dfrac{1}{\left(\frac{1}{p^{6}}\times q^8\right)^{4}} }\)

\(\dfrac{-5\cdot \dfrac{1}{\left(\frac{1}{m^{4}}\times \frac{1}{n^{5}} \right)^{3}} }{7\cdot \dfrac{1}{\left(\frac{1}{p^{6}}\times q^8\right)^{4}} } = \dfrac{-5\cdot {\left(m^4\times n^5\right)^3} }{7\cdot \left( \dfrac{p^6}{q^8\right)^{4}} }\)

\(\dfrac{-5\cdot {\left(m^4\times n^5\right)^3} }{7\cdot \left( \dfrac{p^6}{q^8\right)^{4}} } = \dfrac{-5\cdot {\left(m^{12}\times n^{15}\right)} }{7} \times \left(\dfrac{q^8}{p^6}\right)^4 }\)

\(\dfrac{-5\cdot {\left(m^{12}\times n^{15}\right)} }{7} \times \left(\dfrac{q^8}{p^6}\right)^4 } = \dfrac{-5\cdot {\left(m^{12}\times n^{15}\right)} }{7} \times \left(\dfrac{q^{32}}{p^{24}}\right) }\)

\(\dfrac{-5\cdot {\left(m^{12}\times n^{15}\right)} }{7} \times \left(\dfrac{q^{32}}{p^{24}}\right) } = \dfrac{-5\times {m^{12}\times n^{15}\times q^{32}} }{7\times p^{24}}\)

\(\dfrac{-5\cdot \left(m^{-4}\cdot n^{-5} \right)^{-3}}{7\cdot \left(p^{-6}\cdot q^8\right)^{-4}} = \dfrac{-5\times {m^{12}\times n^{15}\times q^{32}} }{7\times p^{24}}\)

6.) (3·x⁻⁴·y⁻⁵·z⁻²)⁻²

(3·x⁻⁴·y⁻⁵·z⁻²)⁻² = \(3\cdot \left(\dfrac{1}{\left(\dfrac{1}{x^4} \cdot \dfrac{1}{y^5} \cdot \dfrac{1}{z^2} } \right)^2\right)\)

\(3\cdot \left(\dfrac{1}{\left(\dfrac{1}{x^4} \cdot \dfrac{1}{y^5} \cdot \dfrac{1}{z^2} } \right)^2\right) = 3\cdot \left(x^4\cdot y^5\cdot x^2\right)^2\)

3·(x⁴·y⁵·x²)² = 3·x⁸·y¹⁰·x⁴

Therefore;

(3·x⁻⁴·y⁻⁵·z⁻²)⁻² = 3·x⁸·y¹⁰·x⁴

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The transformation of Δ ABC  to ΔADE is a dilation by a scale factor of 1/2 and then 180 degrees rotation about the origin.

We have,

In ΔABC,

The coordinates of each point are:

A = (0, 0)

B = (0, -6)

C = (8, -6)

And,

ΔADE,

A = (0, 0)

D = (0, 3)

E = (4, 3)

Now,

We can see that,

If we use a scale factor 1/2 on each the coordinates of A, B, and C and rotated to 180 degrees about the origin.

We get,

A = (0, 0) = (0, 0)

B = (0, - 6) = (0, -3) = (0, 3)

C = (8, -6) = (4, -3) = (4, 3)

Thus,

The transformation of Δ ABC  to ΔADE is a dilation by a scale factor of 1/2 and then 180 degrees rotation about the origin.

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Answer:

50°C

Step-by-step explanation:

(122°F − 32) × 5/9 = 50°C

The height of balloon is 1.99 miles above the ground.

In the event when the line of sight is directed downhill from the horizontal line, an angle is created with it. Angle of depression is the angle produced between the horizontal line and the observer's line of sight when the object being observed is below the level of the observer.

Let's give the balloon's height above the earth and the distance it is from the first milepost horizontally a name.

We can state the following if the angle of this milepost's depression is 24 degrees:

Tan(24)=\(\frac{h}{b}\)

b = \(\frac{h}{tan(24)}\)

We can state the following for the following milepost as there will be (b+1)miles horizontal distance of and a 20 degree depression at that location.

tan(20)=\(\frac{h}{b+1}\)

b +1  = \(\frac{h}{tan(20)}\)

Therefore, substituting equation 1

\(\frac{h}{tan(24)}\) = \(\frac{h}{tan(20)} -1\)

1 =   \(\frac{h}{tan(20)}\) - \(\frac{h}{tan(24)}\)

1 = h× ( \(\frac{h}{tan(20)}\) - \(\frac{h}{tan(24)}\) )

h = 1.99 miles

Consequently, the height of balloon is 1.99 miles above the ground.

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Answer:

I need help on that too help pls

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

if you can actually see the graph the numbers go up by six.

6, 12, 18 and so on all the way to 6 x 5 which is 30 on the chart then you visualize off of the chart trying 6x6 6x7 6x8 and 6x9 until you get the answer of 54 the money price.  

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable X denotes the number of heads in n tosses of an honest coin (n ≥ 30), then X has an approximately normal distribution with mean, \(\mu=\frac{n}{2}\) and standard deviation, \(\sigma=\frac{\sqrt{n}}{2}\).

Here the number of tosses is, n = 2500.

Since n is too large, i.e. n = 2500 > 30, the random variable X follows a normal distribution.

The mean and standard deviation are:

\(\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25\)

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, μ = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

Answer:

The price fell 40%

Step-by-step explanation:

34/85, then convert it to a percentage

Answer:

30 Bags

Step-by-step explanation:

If Tori were to separate her 6 pound bags of sugar into bags of 1/5 pounds, she'd have 5 bags per pound of sugar. multiply 5 by 6 and you get 30.

Answer:

Step-by-step explanation:

We know that:

Work:

Hence, it is not possible.

Answer:

0.16

Step-by-step explanation:

Its 0.16 because it would be easiest to make 8/50 into 16/100 to make it easier to turn into a decimal.

Answer:-3 over 7

Step-by-step explanation:

rise over run

The area of the region enclosed by the curves y=6x² and y=x²+6 is 8.76 unit square and has been drawn.

The definite integral of a function over a limit represents the area under the curve bounded by the limit as given in integration.

As per the given curves,

y=6x² and y=x²+6

The above curves have been drawn.

The intersection points can be calculated by equating both equations as (-1.095,7.2) and (1.095,7.2).

The area enclosed will be as follows,

Area =  \(\int\limits^{1.095}_{-1.095} {[(x^{2}+6)-6x^{2}]} \, dx\)

A = - 5/3[1.095³ + 1.095³] + 6[1.095 + 1.095]

A =  8.76 unit square.

Hence "8.76 unit squares make up the region that is encompassed by the curves y=6x² and y=x²+6".

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If the commercial banking system actually lends out the maximum amount it is able to lend, the excess reserves will be reduced to zero. This is because there will be no excess reserves held by the commercial banking system after lending out $9 billion.

Given that the required reserve ratio is 30 percent and all figures are in billions, the following table shows the total reserves, excess reserves, and required reserves:

Required reserve ratio 30% Checkable deposits $140Billion Reserves required (30% of checkable deposits)$42Billion Reserves held $51Billion Excess reserves held $9Billion Loans outstanding $109Billion Total Securities $100Billion Total Property $10Billion Total Assets $260Billion Stock shares $130Billion Total liabilities $130Billion Net worth (total assets - total liabilities) $130Billion.

Therefore, the total reserves held is $51 billion, and the reserves required is 30% of $140 billion, which is $42 billion.

This implies that the excess reserves are $9 billion.The maximum amount the commercial banking system can lend out is $51 billion minus $42 billion, which is $9 billion.

This indicates that if the commercial banking system actually lends out the maximum amount it is able to lend, the excess reserves will be reduced to zero. This is because there will be no excess reserves held by the commercial banking system after lending out $9 billion.

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Answer:

\(\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}\)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Algebra II

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Differential Equations

Antiderivatives - Integrals

Integration Constant C

Integration Property [Multiplied Constant]:                                                             \(\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx\)

U-Substitution

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

\(\displaystyle y' = \frac{xy}{2 + x^2}\)

\(\displaystyle y(0) = 2\)

Step 2: Rewrite

Separation of Variables

Step 3: Find General Solution Pt. 1

Integration

Step 4: Identify Variables

Identify variables for u-substitution for 2nd integral.

u = 2 + x²

du = 2xdx

Step 5: Find General Solution Pt. 2

Our general solution is  \(\displaystyle |y| = Ce^{\frac{1}{2} ln|2 + x^2|}\).

Step 6: Find Particular Solution

∴ Our particular solution is  \(\displaystyle y = \sqrt{2}e^{\frac{1}{2} ln|2 + x^2|}\).

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Differential Equations

Book: College Calculus 10e

Answer:

B & D

Step-by-step explanation:

If you do 30x10 it would equal 300

So that wouldn't be it. If you do 49x100 that will equal 4,900 which is correct. If you do 40x600 that would equal 24,000 which is also incorrect. The last one is correct because 400x80 is 32,000

(Your welcome!)

When applying the some rule, we only combine two equal ratios at once. Consequently, we deal with pairs of sides and angles that have equivalent ratios.

When two ratios are equal, a percentage is formed; alternatively, two equal ratios can be said to produce a proportion. When we understand that two ratios are equal, you can write a proportion. The ratios of these two numbers are equal.

When two variables are correlated in a manner that their ratios are equal, this is known as a proportional relationship. In a proportional connection, one variable is always a constant value multiplied by the other, which is another way to think of them. The "constant of proportionality" is the term used to describe that constant. Two angles are said to be complimentary if their sum is 90 degrees. Alternatively put, two angles are said to be complimentary if they combine to make a right angle.

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a. Assuming no additional restrictions, there are  800 possible three-digit area codes.

b. Considering ERCs, there are  80 ERCs.

c. For the seven digits after the area code, there are \(8 \times 10^6 = 8,000,000\)  possible seven-digit phone numbers.

d. Assuming only the 0 and 1 restrictions from parts a and c, the number of possible ten-digit phone numbers is 800 \(\times\) 8,000,000 = 6,400,000,000.

a. For three-digit area codes, the first digit cannot be 0 or 1.

Assuming no additional restrictions, there are 8 possibilities for the first digit (2-9) and 10 possibilities for each of the remaining two digits (0-9). Therefore, the total number of three-digit area codes possible under this plan is \(8 \times 10 \times 10 = 800.\)

b. For an ERC (Easily Recognizable Code), the second and third digits of the area code must be the same, and the first digit cannot be 0 or 1. There are 8 possibilities for the first digit (2-9) and 10 possibilities for the third digit (0-9).

Since the second digit must be the same as the third digit, there is only 1 possibility.

Therefore, the total number of ERCs is \(8 \times 1 \times 10 = 80.\)

c. For the seven digits after the area code, the first digit of the three-digit local prefix cannot be 0 or 1.

There are 8 possibilities for the first digit (2-9) and 10 possibilities for each of the remaining six digits (0-9).

Therefore, the total number of seven-digit phone numbers possible is 8 * \(10\times 10 \times 10 \times 10 \times 10 \times 10 = 8,000,000.\)

d. Assuming only the 0 and 1 restrictions from parts a and c, the number of possible ten-digit phone numbers can be calculated by multiplying the number of possibilities for each digit position.

For the area code (part a), there are 800 possibilities.

For the seven digits after the area code (part c), there are 8,000,000 possibilities.

Therefore, the total number of ten-digit phone numbers possible is 800 * 8,000,000 = 6,400,000,000.

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Answer:

i pretty sure it's 17

Step-by-step explanation:

Answer:

17

Step-by-step explanation:

Answer:

m<c = 20º

Step-by-step explanation:

We use these following definitions:

The sum of an interior angle with its exterior is of 180º.

The sum of the three interval angles of a triangle is 180º.

The measure of the exterior angle at point B is 100°.

This means that the measure of the interior angle is: 180 - 100 = 80º

m∠A is 4 times larger than m∠c

This means that the measure of angle c is x, and the measure of angle a is 4x. The other angle, B, is 80º. So

\(x + 4x + 80 = 180\)

\(5x = 100\)

\(x = \frac{100}{5}\)

\(x = 20\)

The measure of angle C is 20º.