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3 years ago

6

A rectangle has length that is 4 inches longer than its width. It’s perimeter is 36 inches. Which of the following is it’s area?

2 answers:

Scilla [17]3 years ago

5 0

Answer:

C. 77

Step-by-step explanation:

Let l be length and w be the width of the rectangle,

Thus, the perimeter of the rectangle = 2 ( length + width )

= 2 ( l + w )

According to the question,

l - w = 4 ⇒ l = 4 + w

Also, 2(l+w)=36

2(4 + w +w)=36

4 + 2 w = 18

2w = 14

⇒ w = 7

So, the width of the rectangle is 7 inches.

And, length = 4 + 7 = 11 inches

Thus, the area of the rectangle = length × width

= 11 × 7

= 77 in²

Option 'C' is correct.

Pepsi [2]3 years ago

4 0

Width= 7
Length= 11
Area: 11•7=77 (C)

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If you take half my age and add 7, you get my age 13 years ago. How old am I?

mestny [16]

Answer:

You are currently 40 years old

Step-by-step explanation:

Let's say the current age is represented by the variable x.

x = current age

The question says that "if you take half my age and add 7, you get my age 13 years ago"

This can be represented like this:

1/2(x) + 7 = x - 13

Now we solve for x using basic algebra:

1/2(x) = x - 20

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To check if this is correct, plug it back into the equation and see if both sides equal each other:

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2 years ago

How can csc^2-cot^2 = 1?

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ANSWER

By simplifying the left hand side using the Pythagorean Identity.

EXPLANATION

The given identity is

$\csc^{2} (x) - \cot^{2} (x) = 1$

Take the left hand side and simplify to get the right hand side.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1}{\sin^{2} (x)} - \frac{ \cos^{2} (x)}{\sin^{2} (x)}$

Collect LCM for the denominators.

$\csc^{2} (x) - \cot^{2} (x) = \frac{1 - \cos^{2} (x)}{\sin^{2} (x)}$

Recall the Pythagorean Identity.

$\cos^{2} (x) + \sin^{2} (x) = 1$

This implies that:

$1 - \cos^{2} (x) = \sin^{2} (x)$

We substitute this to get,

$\csc^{2} (x) - \cot^{2} (x) = \frac{\sin^{2} (x)}{\sin^{2} (x)}$

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2 years ago

A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

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