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2 years ago

11

Triangle ABC A(-3,1) B(1,3) C(2,-4) . Move point C so that the triangle becomes right angle. Is there a methods apart from guess

and check?

1 answer:

Alex777 [14]2 years ago

7 0

Draw a cartesian graph and draw the points in , then see where c should be to get a right angled triangle

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Solve for x 27^(2x+4)=1/9^(x-46)

lys-0071 [83]

$27^{2x+4}=(\frac{1}{9})^{x-46} \\
(3^3)^{2x+4}=(3^{-2})^{x-46} \\
3^{3(2x+4)}=3^{-2(x-46)} \\
3^{6x+12}=3^{-2x+92} \\
6x+12=-2x+92 \\
6x+2x=92-12 \\
8x=80 \\
x=10$

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2 years ago

Please help me with this 30 points

gogolik [260]

Answer:

D

Step-by-step explanation:

For every 1 unit, it goes vertically 1/2

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Help! I can’t get this!

8_murik_8 [283]

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2 years ago

Read 2 more answers

Given h(x) = –2 – 3, find h(-6).

igomit [66]

Answer

3

Step-by-step explanation:

h(-6)=-(-6)-3

h(-6)=6-3

h(-6)=3

4 0

2 years ago

Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx

shutvik [7]

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$ , so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$ :

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

$\csc v-\cot v=x+C$

$\csc(x+y)-\cot(x+y)=x+C$

Given that $y(0)=\dfrac\pi2$ , we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

$\csc(x+y)-\cot(x+y)=x+1$

5 0

2 years ago