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3 years ago

8

Find the height of the pyramid. Volume

1 answer:

MA_775_DIABLO [31]3 years ago

3 0

Answer:

5

Step-by-step explanation:

like my answer please...

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Help anyone please :)

BARSIC [14]

Nah... The answer is Aida’s work. 5 to the power of 4 is the same as 5x5x5x5.

8 0

4 years ago

Dora is thinking of a number between 500 and 600

wariber [46]

9514 1404 393

Answer:

500 ÷ 8 = 62 r 4, for example

Step-by-step explanation:

In order to have a remainder of 4, the divisor must be greater than 4, so could be any of 5, 6, 7, 8, 9.

The corresponding numbers could be any of ...

504 +5n . . . 0 ≤ n ≤ 19

502 +6n . . . 0 ≤ n ≤ 16

501 +7n . . . 0 ≤ n ≤ 14

500 +8n . . . 0 ≤ n ≤ 12

508 +9n . . . 0 ≤ n ≤ 10

__

For example, with the divisor being 6, and n=13, the number could be ...

580 = 6·96 + 4

7 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

Draw two different perimeters of a rectangle with an area of 14 units. Name their dimensions.

Alenkasestr [34]

1:
Length: 7
Width:2
Perimeter: 18

2:
Length: 5
Width: 2.8
Perimeter: 15.6

8 0

3 years ago

Which is the order of rotational symmetry for a rhombus

Dennis_Churaev [7]

Answer:

Order 2.

Step-by-step explanation:

5 0

3 years ago