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3 years ago

There are 28 students in math class. 22 of them passed the test. What percentage passed the test?

1 answer:

8 0

In order to solve this question, we need to set up a proportion. We are trying to solve for a percentage, so the proportion will be 22/28 = x/100, where x is the percentage that 22 is out of 28.

22 / 28 = x / 100
2200 = 28x Cross multiply
x = 78.6% Divide

22 is about 78.6% of 28. Remember that, whenever you need a percentage, just set up a proportion like this one.

Hope this helps!

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Math question, note: there are two answers!!

stiks02 [169]

The answer is A and B. I showed my work below. I hope it makes sense. Let me know if you have any questions.

5 0

3 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 $L\approx 15.57$ .

8 0

3 years ago

Because of the rainy season, the depth in a pond increases 3% each week. Before the rainy season started, the pond was 10 feet d

Tanzania [10]

I believe it's D
r has to be 1.03

4 0

3 years ago

. Here is an equation 2x + 9 = -15. Write three different equations that have the same solution as 2x + 9 = -15. Show or explain

Anettt [7]

Answer:

1. $x-3=-15$
2. $2x=-24$
3. $2x-3=-27$

Step-by-step explanation:

Given the following question:

$2x+9=-15$

To find three equations that have the same solution as the given equation we have to first solve the equation.

$2x+9=-15$
$9-9=0$
$-15-9=-24$
$2x=-24$
$2x\div2=x$
$-24\div2=-12$
$x=-12$

Equation one:
$x-3=-15$
$-3+3=0$
$-15+3=-12$
$x=-12$

Equation two:
 $2x=-24$
$2x\div2=x$
$-24\div2=-12$
$x=-12$

Equation three:
$2x-3=-27$
$-3+3=0$
$-27+3=-24$
$-24\div2=-12$
$x=-12$

Hope this helps.