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4 years ago

8

Find the surface area of a rectangular prism that has the following dimensions.

2 answers:

guajiro [1.7K]4 years ago

5 0

Look at the pics they will tell you how I got 306 in^2

faltersainse [42]4 years ago

3 0

2(wl+hl+hw)=2·(6·4.5+12·4.5+12·6)=306

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Help with this problem plz!

Tomtit [17]

Use photos math it helps alot

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3 years ago

Y=4mx-7b solve for x

muminat

Your answer is x = 7b/4m + y/4m

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4 years ago

The expression 12x+8 represents the perimeter of a square write 12x+8 as a product then tell what expression represents one side

Pani-rosa [81]

Answer:

Step-by-step explanation:

Find the greatest common factor of 12x + 8

4(3x + 2) so one side is 4 and the other is 3x + 2

5 0

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 $L\approx 15.57$ .

8 0

3 years ago

Algebra 2 Help! Thank you.

pychu [463]

Answer:

-2

Step-by-step explanation:

f(1)=3(1)-5 = 3-5= -2

4 0

3 years ago