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3 years ago

A cube has an edge measuring 4 inches what is the surface area of the cube in inches^2

Mathematics

1 answer:

Arada [10]3 years ago

5 0

Answer:

96^2in

Step-by-step explanation:

4x4=16

16x6=96

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The diameter of a coin is 26mm. Mathew lined up 56 coins from the left end of his desk to the right end of his desk. What is the

Ne4ueva [31]

Answer:

Your answer is 145.6

Step-by-step explanation:

First multiply 56 and 26 which then you get 1456mm then divide the length value by ten and then you get 145.6.

Hope I helped!

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2 years ago

Tom has saved $128, and his goal is to save $500. How much more must he save? Which equation BEST represents this situation?

Jet001 [13]

I think the answer is A

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A group of band instructors answered a survey about hours of rehearsal per week and number of competitions won. The graph shows

IRISSAK [1]

Answer: B. 31

Step-by-step explanation:

This linear regression was constructed by relating the hours practiced per week and the number of competitions won.

Going by this graph, the number of competitions they can expect to win at 5 practices a week is 31.

This is derived by looking for the point where 5 competitions on the x-axis intersects with the line. This point is at 31 competitions on the y axis which would make it the answer.

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3 years ago

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1 feet = _____ inches A.2 inches B.12 inches C.1.2 inches D.1 inch

Advocard [28]

Answer:

its B!!!!!!

Step-by-step explanation:

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3 years ago

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Use cylindrical coordinates. Find the volume of the solid that lies within both the cylinder x2 + y2 = 25 and the sphere x2 + y2

Inessa05 [86]

Answer:

Step-by-step explanation:

we are asked to find the volume of solid that lies within both the cylinder

$x^2 + y^2 = 25$

and

the sphere $x^2 + y^2 + z^2 = 49.$

Conversion from rectangular to cylindrical is

$x=rcost\\y = rsint\\z=z$

|J| =r

In cylindrical coordinates the volume is bounded by the cylinder r=5 and

$r^2+z^2 =49$

Hence we can write volume as

$\int \int \int dxdydz\\=\\\int _0^5 \int_0^{2\pi} \int_{-\sqrt{49-r^2} } ^{\sqrt{49-r^2} rdzdtdr\\= 2\pi \int _0^5 (2\sqrt{49-r^2} rdr\\=4\pi (-(49-r^2) (2/3)\\= \frac{4\pi}{3} (343-48\sqrt{6} )$

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3 years ago