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3 years ago

5

Rotation 90° clockwise about the origin

1 answer:

lora16 [44]3 years ago

6 0

Answer:

Take the picture you uploaded.

Click the rotate 'button' once.

Change the x to y and y to x on the graph. (axis labels)

Done

J (0, -1)

K(-4,-3)

I ( -4,-1)

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Please help!!! It's due rn!

nexus9112 [7]

Step-by-step explanation:

In triangles BAD and BCD ,

BD=BD (common)

angle BDA= angle BCD {90°each(given)}

AD=DC (given)

.•. traingle BAD is congruent to triangle BCD (SAS criterion)

Hence , angle A = angle C (CPCT)

8 0

2 years ago

All the edges of a regular square pyramid have length 8 . What is the:

user100 [1]

Answer:

vol=BX1/3 h

=64* 1/3 square root of 32

Step-by-step explanation:

5 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

12 divided by 4y+6 = 2/7

Mila [183]

Answer:

The correct answer is y = 9

Step-by-step explanation:

It is given that,

12/(4y + 6) = 2/7

<u>To find the value of y</u>

12/(4y + 6) = 2/7

By cross multiplying,

12 * 7 = 2 * (4y + 6 )

84 = 8y + 12

8y + 12 = 84

8y = 84 - 12

8y = 72

y = 72/8 = 9

y = 9

Therefore by solving the given expression we get the value of y = 9

3 0

3 years ago

Read 2 more answers

Solve the following equation |x – 3| = 4. Show all work.

Igoryamba

<h2>Solving Equations with Absolute Values</h2><h3>Answer:</h3>

$x = 7$ and $x = -1$

<h3>Step-by-step explanation:</h3>

Solving for the Positive Absolute Value:

$|x -3| = 4 \\ x -3 = 4 \\ x -3 +3 = 4 +3 \\ x = 7$

Solving for the Negative Absolute Value:

$|x -3| = 4 \\ -(x -3) = 4 \\ -(x -3) \cdot -1 = 4 \cdot -1 \\ x -3 = -4 \\ x -3 +3 = -4 +3 \\ x = -1$

8 0

3 years ago

Read 2 more answers