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2 years ago

4, 16, 13, 52, 49, 196, 193, 772, _ What comes next?

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

4 0

12
3
39
3
147
3
579
Just subtract each number

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What is the volume in cubic centimeters of a rectangular prism that has a length of 6.2 centimeters, a width of 3.5 centimeters,

Rufina [12.5K]

The length is 6.2
the width is 3.5
the height is 10
So the volume =217cm^3

I hope this help

5 0

2 years ago

Read 2 more answers

Explain hovevou could estimate 22% of 78 to check if your answer is reasonable.

fomenos

Answer:

78 rounds to 80, so find 10% of 80 and double it. 10% of 80 is 8. So, 20% of 80 is 16. Then, 22% of 78 is a little more than 16.

78 rounds to 80, and 22% is a little less than 25%. 25% (or one-fourth) of 80 is 20. So, 22% of 78 is a little less than 20.

Step-by-step explanation:

8 0

2 years ago

Does 12, 24, 36 form a right triangle

gogolik [260]

Explanation is in the file

tinyurl.com/wpazsebu

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2 years ago

x+(x+3)+x^2=75 Please help........ Everyone is giving me answers i know arent true...

son4ous [18]

x + (x + 3) + $x^{2}$ = 75

so you will add up all of the x 's but NOT the x^2

2x + 3 + $x^{2}$ = 75

Rearrange to put the x^2 in front of 2x

$x^{2}$ + 2x + 3 = 75

subtract the 75 from both sides

$x^{2}$ + 2x - 72 = 0

8 0

2 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

2 years ago