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4 years ago

Which is a right triangle formed using a diagonal through the interior of the cube?

Mathematics

2 answers:

QveST [7]4 years ago

7 0

The answer is in fact, A. or "ABE"

charle [14.2K]4 years ago

4 0

Draw the given answers to determine the correct solution.

ΔABE is correct because it is a right triangle through a diagonal
ΔABD is incorrect because it does not go through diagonals.
ΔADH is incorrect because it does not go through diagonals.
ΔACE is incorrect because it does not go through the interior of the cube.

Answer: ΔABE

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Given a 12 inches by 12 inches square has a 3 inches by 3 inches square cut from the center, what is the remaining area? Please

devlian [24]

The answer is the area of the large square - the area of the cut out square

Area of large square: 12*12 = 144

Area of cut out square: 3*3 = 9

144 - 9 = 135

Answer: 135 square inches

6 0

3 years ago

The area of the rectangle is 4x2, what does the coefficient 4 mean in terms of the problem?

12345 [234]

Width is x
length is 4x

answer b is correct the length is 4 times the width

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4 years ago

when 9 is increased by 3x the result is greater than 36. What is the least possible interger value for x?

arsen [322]

Answer:

Step-by-step explanation:

"When 9 is increased by 3x"

THis means 9 + 3x

"The result is greater than 36"

We use greater than sign and 36

So we can write:

9 + 3x > 36

SOlving this via equation rules and algebra:

9 + 3x > 36

3x > 36 - 9

3x > 27

x > 27/3

x > 9

This means x is everything greater than 9, which satisfies the equation. So in terms of integers, it can be 10, 11, 12, 13...anything above

We want to find the least possible integer value of x, so it is definitely 10

5 0

4 years ago

Read 2 more answers

Find the definite integral from 0 to 1/16 of arcsin(8x)/sqrt(1-64x^2)dx

kolezko [41]

$\displaystyle\int_0^{1/16}\frac{\arcsin8x}{\sqrt{1-64x^2}}\,\mathrm dx$

First let $y=8x$ , so that $\mathrm dx=\dfrac{\mathrm dy}8$ to write the integral as

$\displaystyle\frac18\int_0^{1/2}\frac{\arcsin y}{\sqrt{1-y^2}}\,\mathrm dy$

Now recall that $(\arcsin y)'=\dfrac1{\sqrt{1-y^2}}$ , so substituting $z=\arcsin y$ should do the trick. The integral then becomes

$\displaystyle\frac18\int_0^{\pi/6}z\,\mathrm dz=\frac1{16}z^2\bigg|_{z=0}^{z=\pi/6}=\frac{\pi^2}{576}$

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3 years ago

State true or false for each statement

Neko [114]

Answer:

true

false

Step-by-step explanation:

I think that that is right mark brainliest if i'm right

4 0

3 years ago