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2 years ago

Need the (x) & (y) value

Mathematics

1 answer:

Katen [24]2 years ago

3 0

Answer:

x=0

y=6

Step-by-step explanation:

Two equations with two unknowns can be solved. Use the three steps given below.

In general you can save time if you use the simplest expression for step 1. You can choose either expression to start with x =.. or you can rewrite it, so it starts with y =...

FOLLOW THESE STEPS:

1. Rewrite the first expression so it starts with y =...

2. Then use that expression and substitute it in the other expression, to find the value for x.

3. Use that value found in step 2. and substitute it in the rewritten expssion you created in step 1. Solve this last expression and you'll get the value for y.

Step 1.

-6y + 11x = -36

-6y = -11x -36

multiply left and right of the = sign by -1

6y = 11x + 36

divide left and right of the = sign by 6

y = 11x/6 + 36/6

y = 11/6x + 6

Step 2.

-4 * ( y ) + 7x = -24

-4 * ( 11/6x + 6 ) + 7x = -24

( -44/6x - 24 ) + 7x = -24

-44/6x + 7x = -24 + 24

You can see immediately that ...x = 0

Just for fun of it ....

-44/6x + 42/6x = 24 - 24

(-44+42)/6x = 0

-2/6x = 0

-1/3x = 0

multiply left and right of the = sign by -3

3/3x = 0 * -3

x = 0

Step 3.

y = 11/6( x ) + 6. and substitute x = 0 gives:

y = 11/6( 0) + 6

y = 0 + 6

y = 6

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Dmitry [639]

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4 0

3 years ago

Express the equation in rectangular form<br> a) r = 2sin(2Θ)<br> b) r = 4cos(3Θ)

Dafna11 [192]

Answer:

a. sqrt(x^2+y^2) * (x^2+y^2) = 4xy

b. (x^2+y^2)^2 = 4x{ (x^2 - 3 y^2}

Step-by-step explanation:

a) r = 2sin(2Θ)

r* = 2 sin 2 theta

We know sin 2theta = 2 sin theta cos theta

r = 2 * 2 sin theta cos theta

sin theta = y/r and cos theta = x/r

r = 4 y/r * x/r

r = 4xy / r^2

Multiply each side by r^2

r^3 = 4xy

r * r^2 = 4xy

We know r = sqrt(x^2+y^2) and r^2 = (x^2+y^2)

sqrt(x^2+y^2) * (x^2+y^2) = 4xy

b) r = 4cos(3Θ)

we know that cos 3 theta = cos^3(theta) - 3 sin^2(theta) cos(theta)

r = 4 *cos^3(theta) - 3 sin^2(theta) cos(theta)

Factor out a cos theta

r = 4 *cos(theta){ cos^2 (theta) - 3 sin^2(theta) }

We know that sin theta = y/r and cos theta = x/r

r = 4 (x/r) { (x/r)^2 - 3 (y/r)^2}

r = 4 *(x/r) { (x^2/r^2 - 3 y^2/r^2}

Multiply by r

r^2 = 4x { (x^2/r^2 - 3 y^2/r^2}

Multiply by r^2

r^2 *r^2 = 4x { (x^2 - 3 y^2}

r^4 = 4x{ (x^2 - 3 y^2}

We know r^2 = (x^2+y^2)

(x^2+y^2)^2 = 4x{ (x^2 - 3 y^2}

5 0

3 years ago

What is the solution to the equation -2(1 + 5a) = -6(a + 3) <br> a=

Yuliya22 [10]

-2(1+5a)=-6(a+3)
Expand the brackets:
-2-10a=-6a-18
Add 10a to each side:
-2=4a-18
Add 18 to each side:
16=4a
Divide each side by 4:
4=a

Hope this helps :)

6 0

3 years ago

Seth took 270 pictures for the yearbook And 180 for the school paper. Which proportion can be used to determine p,the percent of

seraphim [82]

F because, if you take 270/450=.6 or 60% and that would be your p, and 60%/100=.6

Enjoy!=)

3 0

3 years ago

Read 2 more answers

Calculus 2 master needed; evaluate the integral PLEASE SHOW STEPS IF IM WRONG <img src="https://tex.z-dn.net/?f=%5Cint%7Bsin%5E3

sweet [91]

Answer:

Yes, you answer is correct! It just needs to be simplified :)

Step-by-step explanation:

So we have the integral:

$\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx$

As you had done, we can split off the numerator:

$=\int \frac{\sin(x)(\sin^2(x))}{\sqrt{\cos(x)}}dx$

Using the Pythagorean Identity, this is:

$=\int \frac{\sin(x)(1-\cos^2(x))}{\sqrt{\cos(x)}}dx$

Now, we can do u-substitution. Let u equal cos(x). Thus:

$u=\cos(x)\\du=-\sin(x)dx\\-du=\sin(x)dx$

So:

$=\int \frac{1-u^2}{\sqrt{u}}(-du)$

Simplify:

$=-\int\frac{1-u^2}{\sqrt u}du$

We can then split the terms:

$=-\int \frac{1}{\sqrt u}-\frac{u^2}{\sqrt u}du$

Expand the integral:

$=-(\int \frac{1}{\sqrt u}du-\int\frac{u^2}{\sqrt u}du)$

Simplify each of the u.

For the left, that is simply u^-1/2.

For the right, it is u^(2-1/2) or u^3/2. Thus:

$=-(\int u^{-\frac{1}{2}}du-\int u^{\frac{3}{2}}du)$

Reverse Power Rule:

$=-(\frac{u^{1+-\frac{1}{2}}}{1+-\frac{1}{2}}-\frac{u^{1+\frac{3}{2}}}{1+\frac{3}{2}})$

Simplify:

$=-(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{5}{2}}}{\frac{5}{2}})$

Simplify further:

$=-(2u^{\frac{1}{2}}-\frac{2u^{\frac{5}{2}}}{5})$

Distribute the negative:

$=-2u^{\frac{1}{2}}+\frac{2u^{\frac{5}{2}}}{5}$

And substitute back cos(x) for u:

$=-2\cos^{\frac{1}{2}}(x)+\frac{2\cos^{\frac{5}{2}}(x)}{5}$

And this is precisely what you got, so well done!

We can simplify this by first multiplying the first term by 5 to get a common denominator. So:

$=-\frac{10\cos^{\frac{1}{2}}(x)}{5}+\frac{2\cos^{\frac{5}{2}}(x)}{5}$

Combine:

$=\frac{-10\cos^{\frac{1}{2}}(x)+2\cos^{\frac{5}{2}}(x)}{5}$

Factor out a cos^(1/2)(x) and a 2. Since we factored out a cos^(1/2)(x), we need to subtract their exponents inside. Thus:

$=\frac{2\cos^{\frac{1}{2}}(x)(-5\cos^{\frac{1}{2}-\frac{1}{2}}(x)+\cos^{\frac{5}{2}-\frac{1}{2}}(x))}{5}$

Simplify:

$=\frac{2\cos^{\frac{1}{2}}(x)(-5+\cos^2(x))}{5}$

Simplify:

$=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}$

And, of course, C:

$=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C$

So:

$\int \frac{\sin^3(x)}{\sqrt{\cos(x)}}dx=\frac{2\sqrt{\cos{x}}(\cos^2(x)-5)}{5}+C$

And we're done :)

3 0

3 years ago

Read 2 more answers