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Ray Of Light [21]

3 years ago

12

20x-4y=36 solve for y need help plz

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

5x-9

Step-by-step explanation:

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Make y the subject: <br><br> 5x + 2y = 11

Anton [14]

Does that mean solve for y? If so, here it is:
5x + 2y = 11.

Move the x term by subtracting.
2y = -5x + 11

Divide by 2.
y = -5/2x + 11/2

7 0

3 years ago

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How do you solve: 3In5+4Inx

Dafna11 [192]

Answer:

ln(125x $x^{4}$ )

Step-by-step explanation:

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3 years ago

The slope is 35. The y-intercept is 550. what can you conclude?

MrRa [10]

Answer:

the formula is y = mx + b. (m) represents slope and (b) represents the intercept

Step-by-step explanation:

So with that formula and info given, you can form the equation y = 35x + 550

6 0

4 years ago

Find the square root of 167281 by division method

tensa zangetsu [6.8K]

Answer:

square root of 167281 is 409

Here's the Step-by-step explanation using division:

6 0

3 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago