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Mkey [24]
3 years ago
15

What is the value of the number 2 if it is in the ten-thousands place?

Mathematics
1 answer:
brilliants [131]3 years ago
8 0

Answer:

contiua seno d2n

Step-by-step explanation:

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The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance wo
Keith_Richards [23]
This question has this set of answer choices:

a) when the plane cuts three faces of the cube, separating one corner from the others

b) when the plane passes through a pair of vertices that do not share a common face

c) when the plane is perpendicular to the base and intersects two adjacent vertical faces

d) when the plane makes an acute angle to the base and intersects three vertical faces

e) not enough information to answer the question

The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others

You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.

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3 years ago
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Rewrite the expression without using grouping symbols. 4(x – 2)
tangare [24]

out of these, it would be 4x - 4 * 2

7 0
2 years ago
Please help me with this...it’s 2am and I still have 3 more assignments...thank you
BARSIC [14]

Answer:

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
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