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2 years ago

The formula for the perimeter of a rectangle is P=2(l+w).

Mathematics

1 answer:

olga2289 [7]2 years ago

7 0

Answer:

See below.

Step-by-step explanation:

Part A.

I'm changing the l and w to capital letters, so you don't confuse the l (small-case L) with a 1 (one).

P =2(L + W)

Since W is on the right side, switch sides to end up with W on the left side.

2(L + W) = P

Since the quantity in parentheses is being multiplied by 2, we divide both sides by 2.

L + W = P/2

L is being added to W, so we subtract L from both sides.

W = P/2 - L

Part B.

First, I switched sides to have W on the left side. Then, I divided both sides by 2. Finally, I subtracted L from both sides.

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Write the next three terms of the geometric sequence.

Sati [7]

Answer:

5, 1, 0.2

Step-by-step explanation:

Divide the number by 5 to get the next one.

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3 years ago

A candy bar is rectangular in shape and measures 3 inches by 4 inches by .5 inches. How many of the candy bars will fit into the

allsm [11]

It will fit 17 candy bars. Hope this helps!

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3 years ago

The student council had six fund raisers during the school year. They raised $212, $205, $206, $190, $188, and $195. Using the c

Rus_ich [418]

Answer:

$1200

Step-by-step explanation:

The clustering method for estimation is a method used for estimating the sums and products of numbers. This method is used when the numbers that are to be multiplied or added are near in value to a single number.

Given the money raised by six fund raisers:

$212, $205, $206, $190, $188, and $195

The numbers are to be added to find the total money raised. i.e.:

$212 + $205 + $206 + $190 + $188 + $195

But all the numbers are close to $200. Hence:

Using clustering method = $200 + $200 + $200 + $200 + $200 + $200 = 6 * $200 = $1200

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2 years ago

You have 87 marbles <br><br> Round that number to the nearest ten.

VashaNatasha [74]

Answer: 90

Step-by-step explanation:

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2 years ago

Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

ladessa [460]

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

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2 years ago