Ask question

Ask question

All categories

3 years ago

7

Pic below help ASAP (not exam)

1 answer:

mel-nik [20]3 years ago

6 0

Divide 9.3 by 3.1:

9.3 / 3.1 = 3

subtract the exponents: 34 -17 = 17

then turn those into scientific notation:

3 x 10^17

You might be interested in

HELP ME PLEASE PLEASE IM BEGGING

suter [353]

Answer:

FALSE, (2, 9) is not a solution to the set of inequalities given.

Step-by-step explanation:

Simply replace x by 2 and y by 9 in the inequalities and see if the inequality is true or not:

irst inequality:

$y\geq 4x\\9\geq 4\,(2)\\9\geq 8$

so thi inequality is verified as true since 9 is larger or equal than 8

Now the second inequality:

$y$

This is FALSE since 9 is larger than 4 (not smaller)

Therefore the answer to the question is FALSE, (2, 9) is not a solution to the set of inequalities given.

6 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

is this right?? plz answer

tekilochka [14]

Answer:maybe

Step-by-step explanation:

To answer that we need to talk about parallel universes

8 0

3 years ago

Read 2 more answers

How does the length of the hypotenuse ina right triangle relate to the lengths of the legs?

Simora [160]

I am pretty sure it is "D"

7 0

3 years ago

10/5 POINTSSSSSS!!!!!

Yanka [14]

It's b 1/9 mile i thank

8 0

3 years ago