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1 year ago

If AC = 8x - 14 and EC = 2x + 11, solve for x.

Mathematics

1 answer:

Elan Coil [88]1 year ago

5 0

Answer:

Step-by-step explanation:

8x-14=2x+11

-8 -8

-14=-6x+11

-11

-25=-6x

/-6

x=4.16

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James had $200 he wanted to buy a pen which cost $5 how many pens can he buy?

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200/5=40
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3 years ago

How did I get a 2 here (simplifying ratios)

PilotLPTM [1.2K]

Answer:

$\frac{2x {}^{2} + 14x + 24 }{4 {x}^{2} + 4x - 48 } = \frac{x + 4}{2(x - 2)}$

Step-by-step explanation:

Please see attached picture for full solution.

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3 years ago

The middle number of 3 consecutive whole numbers is 2a.

TiliK225 [7]

First number --- (2a-1)
middle number ---- (2a)
third number -----(2a+1)
2a-1+2a+2a+1=250
6a=250
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3 years ago

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.

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2 years ago

X-1=y what is the answer plz give me the answer

guajiro [1.7K]

Answer:

First of all, observe that this is a line, since you can rewrite the equation as y−x+1=0 , which follows the pattern of the generic line ax+by+c=0 .

To draw a line you need two of its points, which you can connect.

To find two points, you can plug any value for one variable, and solve for the other.

For example, let's choose

x=0 . The equation becomes

0−1=y , and so y=−1 .

The first point is thus (0,−1).

Then, let's choose

x=1 . The equation becomes 1−1=y , and so y=0 . The first point is thus

(1,0).

Now you only need to draw the points

(0,−1)

Step-by-step explanation:

Hope it is helpful...

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2 years ago