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3 years ago

320 million written out

Mathematics

2 answers:

miv72 [106K]3 years ago

7 0

320,000,000 I think lol

bija089 [108]3 years ago

3 0

32000 just count out the digits

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A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerl

STatiana [176]

A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerleaders make signs similar to the banner. the length of a sign is 11 inches. First solve the width of the rectangle:

1496 sq in/ 44 = 34 in

So the sign has also a width of 34 in and a length of 11 so the area is 34*11 =374 sq in

The perimeter is (34*2) +(11*2) =90 in

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3 years ago

The product of the roots of 5 - 2m - 3m^2 = 0 is: A. -5/3 B. -3/5 C. 5/3

dimaraw [331]

m= −5/3 ,swnkadkjkkjmd,m,mdas

7 0

3 years ago

Read 2 more answers

Plzzzzzzzzzzzzzzzzzzzzzzzzz

AleksandrR [38]

Answer:

i dont know what it means by expand but i will simplify it

it would simplify to 16x-18

Step-by-step explanation:

5(2x-1) + 2(3x-6)

10x-5+6x-12

10x+6x-5-12

16x-18

7 0

3 years ago

A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la

SOVA2 [1]

Answer:

$8\pi\text{ square cm}$

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

$S=2\pi rh$

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

$16 = 4r^2 + h^2$

$16 - 4r^2 = h^2$

$\implies h=\sqrt{16-4r^2}$

Thus, the surface area of the cylinder,

$S=2\pi r(\sqrt{16-4r^2})$

Differentiating with respect to r,

$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

4 0

3 years ago

If w and are numbers, which of the following expressions will always be the distance between w and Q

aev [14]

<h3>Answer: Choice A. |w - q|</h3>

Let's say for example that w = 10 and q = 7. This means the distance between these values is w-q = 10-7 = 3. This is the distance between w and q.

Now let's make q larger. If w = 12 and q = 20, then w-q = 12-20 = -8 assuming we subtract in the same order. We use absolute value bars to ensure the result is positive. So instead we say

|w - q| = |12 - 20| = | -8 | = 8

Distance is never negative.

3 0

3 years ago