3 years ago

(2,-4), (x,\ y), A, and B all lie on the line. Find an equation relating x and y.

Mathematics

1 answer:

Olin [163]3 years ago

4 0

Answer:

I can't see the picture, sorry. :(

Step-by-step explanation:

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Simplify the expression where possible. (3b 3) 4 .

goblinko [34]

Answer: 36b

Step-by-step explanation:

3•3=9 add the 9

9b • 4

8 0

3 years ago

15 points please IF NOT YOUR COMMENT WOULD BE REPORTED ALSO YOUR ACCOUNT pls thank you

olga55 [171]

the answer to this question is C

6 0

3 years ago

KX equals X +4 find K7

Andre45 [30]

KX = x+4 so solve for k.
Divide both sides of the equation by x and you get K=4.
Plug it in, and you get (4)(7)=28
Your answer is 28.

5 0

3 years ago

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Oksana_A [137]

$\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}$

The length of the curve is given by the integral

$\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt$

Expand and rewrite the integrand:

$(e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}$
$=e^{-2t}(e^{4t}+2e^{2t}+1)$
$=e^{-2t}(e^{2t}+1)^2$
$\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}$

Now the integral is

$\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt$
$=2\displaystyle\int_0^3\cosh t\,\mathrm dt$
$=2\sinh t\bigg|_{t=0}^{t=3}$
$=2\sinh 3$

7 0

3 years ago

Where can the prpendicular bisectors of an acute triangle intersect?

ch4aika [34]

The meeting point of the 3 perpendicular bisectors of a triangle's sides is called the circumcenter.
The circumcenter of an acute triangle is always inside the triangle.
so the answer is "a"

Source:
http://www.1728.org/trictr.htm

8 0

4 years ago