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3 years ago

I don't understand how to do it

Mathematics

1 answer:

leonid [27]3 years ago

5 0

I don't know what question you are but basically to find the the area of something u would either add and then multiply or just multiply . It depends on how your teacher taught you

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What is the slope of a line is parallel to y=(1/2)x=3

aniked [119]

Hi there! The answer is (1/2)

I suppose you mean the slope of a line that is parallel to y = (1/2)x + 3.

This equation is written in slope-intercept form, which means that the number in front of x represents the slope of the line. The other number represents the y-intercept.

A line that is parallel to another line, will be a line with the same slope. Since the slope of y = (1/2)x + 3 is (1/2), the slope of a line parallel to this line is (1/2) as well.

~ Hope this helps you!

8 0

3 years ago

At a local school supply store, a highlighter costs $1.25, a ballpoint pen costs $0.80, and a spiral notebook costs $2.75. Expla

egoroff_w [7]

Since the communative property shows you can add or multipy in any order, this means you can add all three of these in a different order and still get the same answer.....For example, you can do $1.25 + $0.80 + $2.75 or you can switch it around and do it in another way

4 0

3 years ago

6 points and brainliest answer

Damm [24]

If m<2 is 18 then m<1 is 72

8 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLISET!!! Drag and drop the answers into the boxes to identify the vertex and axis of symmetry of this function.

tensa zangetsu [6.8K]

Answer:

2 -1 -2?..

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

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 y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 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 x 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 y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago