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3 years ago

12

I need help this is due today

2 answers:

Jet001 [13]3 years ago

5 0

Answer:

Step-by-step explanation:

78:33

Lana71 [14]3 years ago

4 0

Divide actual distance by the distance of the scale miles

39 miles / 13 miles = 3

Multiply that by the scale inches:

2 inches x 3 = 6 inches.

The answer is 6 inches.

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PLEASE HELP ME PLEASE

suter [353]

Answer:

$1.21

Step-by-step explanation:

Start by finding the total amount spent on shirts. Do this by multiplying $3.65 and 4 together. You will get 14.6. So, he spent $14.60 on 4 shirts. Take this amount and subtract it from the total amount paid for shirts and socks. $23.07 minus $14.60. This will give you $8.47 left to spent on socks. $8.47 divided by 7 is $1.21. Therfore, James paid $1.21 per pair of socks.

4 0

3 years ago

6 times the sum of 12 and 8

White raven [17]

12 * 8 = 96 * 6 = 576

6 0

3 years ago

Read 2 more answers

Evaluate: 5-3 xwhen x=-7 Substitute-7 for x.

Mice21 [21]

The answer for this question is 26

7 0

3 years ago

PLEASE HELP I WILL GIVE BRAINLIEST

Triss [41]

Answer:

20 inches²

Step-by-step explanation:

perimeter = 8 + 3 + 5 + 4

= 20 inches²

4 0

2 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

2 years ago