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3 years ago

8

What would F^-1(-2) equal

1 answer:

Reil [10]3 years ago

4 0

Answer:

I hope this helps:)

Step-by-step explanation:

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What equation is perpendicular to 4y=3x-8 and passes through (-12,7)

-BARSIC- [3]

Here is the answer to your problem

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3 years ago

Hiii please help i’ll give brainliest:)

Anit [1.1K]

Answer:

Disjointed

Step-by-step explanation:

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6 0

3 years ago

Help me please..................

Rudiy27

253. First, I squared 8, to get 64, then multiplied that by 12. I subtracted 9 from this because 3x3 is 9 to get 759. Then, divide this by 3 to get 253. Hope this helped!

8 0

3 years ago

Read 2 more answers

Solve for x {x+4]= 6

wlad13 [49]

The solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

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

We have an equation:

x(x + 4) = 6

By distributive property:

x² + 4x = 6

x² + 4x - 6 = 0

a = 1, b = 4, c = -6

Plugging all the values in the formula:

$\rm x = \dfrac{-4 \pm\sqrt{4^2-4(1)(-6)}}{2(1)}$

After calculating:

$\rm x = \dfrac{-4 \pm\sqrt{40}}{2}$

$\rm x=-2+\sqrt{10},\ \ \ or \ \ \ x=-2-\sqrt{10}$

Thus, the solution to the equation x(x+4) = 6 is x = -2 + √10 or x = -2 - √10 after solving with the quadratic formula.

Learn more about quadratic equations here:

brainly.com/question/2263981

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6 0

2 years ago

Read 2 more answers

If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)

sweet-ann [11.9K]

Answer:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

Step-by-step explanation:

Given

$tan\theta = 3$

Required

Find

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$

Calculate $tan(\theta + \pi) \ and\ tan(\theta+2\pi)$

Using tan rule

$tan(\theta + \pi) = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}$

So:

$tan(\theta + \pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + \pi) = \frac{tan\theta}{1 }$

$tan(\theta + \pi) = \tan\theta$

$tan(\theta+2\pi)$

$tan(\theta + 2\pi) = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}$

$tan(\theta + 2\pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + 2\pi) = \tan\theta$ '

'So:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$ $= tan\theta +tan\theta +tan\theta$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 3+3+3$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

4 0

2 years ago