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3 years ago

What is x^2 + v = w when you have to make x the subjectim really stuck

Mathematics

2 answers:

ipn [44]3 years ago

6 0

X²+v=w (move v to w)
x²=w-v
x=√w-v (then square root w-v)

ArbitrLikvidat [17]3 years ago

5 0

Substract v to both sides of the equation: x^2+v-v = w-v
you get x^2=w-v
Now take the square root of both sides: root(x^2) = root(w-v)
you get x= sq root (w-v) or x= +/- sq root(w-v)

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3 years ago

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For each ordered pair, determine whether it is a solution to the system of equations.

-Dominant- [34]

The only ordered pair that is a solution to the given system of equations is (-2, -6)

<h3>System of Linear Equations </h3>

From the question, we are to determine if each ordered pair is a solution to the given system of equations

The given system of equations is

-9x + 2y = 6

5x - 3y = 8

For (7, 9)

That is,

x = 7, y = 9

Putting the values into the first equation

Is -9(7) + 2(9) = 6

-63 + 18 = 6

-45 ≠ 6

Thus, (7,9) is not a solution

For (0, 3)

That is,

x = 0, y = 3

Putting the values into the first equation

Is -9(0) + 2(3) = 6

0 + 6 = 6

6 = 6

The ordered pair satisfies the first equation

Testing for the second equation

Is 5(0) - 3(3) = 8

0 - 9 = 8

-9 ≠ 8

Thus, (0, 3) is not a solution

For (5, -4)

That is,

x = 5, y = -4

Putting the values into the first equation

Is -9(5) + 2(-4) = 6

-45 - 8 = 6

-53 ≠ 6

Thus, (-5,4) is not a solution

For (-2, -6)

That is,

x = -2, y = -6

Putting the values into the first equation

Is -9(-2) + 2(-6) = 6

18 - 12 = 6

6 = 6

The ordered pair satisfies the first equation

Testing for the second equation

Is 5(-2) -3(-6) = 8

-10 + 18 = 8

8 = 8

The ordered pair satisfies the second equation

∴ The ordered pair that is a solution to the system of equations is (-2, -6)

Hence, the only ordered pair that is a solution to the given system of equations is (-2, -6)

Learn more on System of equations here: brainly.com/question/8630769

#SPJ1

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1 year ago

Topic: Area of non shaded area.

Dimas [21]

Answer:B) 150

Step-by-step explanation:

Area of rec= L×W

= 10×30= 300

Area of shaded area= 1/2 Base× Height

1/2×10×15=75

75×2 = 150

Area of unshaded part= 300-150= 150

8 0

3 years ago

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I am so confused! Can anyone help me with this trig problem?

Dominik [7]

$\frac{5\pi }{4}$ or $\frac{7\pi }{4}$

divide both sides by 2

sinΘ = - $\frac{\sqrt{2} }{2}$

Since the sin ratio is negative then Θ is in third / fourth quadrants

Θ = $sin^{- 1}$ ( $\frac{\sqrt{2} }{2}$ ) = $\frac{\pi }{4}$ related acute angle

Θ = ( π + $\frac{\pi }{4}$ ) = $\frac{5\pi }{4}$

or Θ = ( 2π - $\frac{\pi }{4}$ ) = $\frac{7\pi }{4}$

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