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3 years ago

What is the value of x when h(x)=-3? -7

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:

○ -1

Step-by-step explanation:

Looking closely at this piecewise function, when the line on the left-hand side intersects at -3 = y, x is -1.

I hope this helps, and as always, I am joyous to assist anyone at any time.

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50 PONITS This model represents an equation. (a) Write an equation that could be used to solve x. (b) Solve the equation for x.

Wittaler [7]

Answer:

A) $5x-1=-6$

B) $x=-1$

Step-by-step explanation:

On the left, we have five (positive) Xs and one -1. So, we can represent the left with the expression $5x-1$ .

On the right, we have six -1s. So, we can represent the right with the expression $-6$ .

Since the balance is balanced, we know that the two expressions are equal to each other. So, we can write the equation:

$5x-1=-6$

Now, we can solve the equation for x. Add 1 to both sides:

$(5x-1)+1=(-6)+1$

The left will cancel. Add on the right:

$5x=-5$

Finally, divide both sides by 5:

$(5x)/5=(-5)/5$

The left side will cancel. So, the value of x is:

$x=-1$

8 0

4 years ago

Solve the equation 8 minus 2X equals 8X +14

lubasha [3.4K]

Answer:

$8 - 2x = 4x + 14 \\ 6x = - 6 \\ x = - \frac{ 6}{6} \\ \therefore \: x = - 1$

7 0

3 years ago

In how many ways can a committee of

Yuri [45]

Answer:

This might be permutation promblem srry if im wrong.

P(6,3)=6x5x4=120

P(4,2)=4x3=12

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3 years ago

Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$