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4 years ago

Relations and functions need help please.

Mathematics

1 answer:

Ierofanga [76]4 years ago

4 0

Just substitute the value of x back into the function.

So, $f(x) = -x^2 + 1$ becomes $f(-3) = -(-3)^2 + 1$

$f(-3) = 3^2 + 1 \\ \\ f(-3) = 9 + 1 \\ \\ f(-3) = 10$

So, f(-3) is equal to 10.

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Tyler had $65.40 in his checking account. Then he wrote checks on the account for $20.53, $13.48, and $19.40. What is the balanc

vesna_86 [32]

$65.40 - $20.53 - $13.48 - $19.40=$11.99

7 0

4 years ago

Dionne can fold 175 packing boxes in 50 mìnutes. Elias can fold 120 packing boxes in 40 minutes. How long will it take each pers

Gre4nikov [31]

<h2>Dionne will fold 210 packing boxes in "60 mìnutes" and</h2><h2>Elias will fold 210 packing boxes in "70 mìnutes".</h2>

Step-by-step explanation:

Given,

Dionne can fold 175 packing boxes = 50 mìnute and

Elias can fold 120 packing boxes = 40 minutes

To find, the total time each person to fold 210 packing boxes = ?

∵ Dionne can fold 175 packing boxes = 50 mìnute

∴ In 1 mìnute, Dionne can fold number of packing boxes = $\dfrac{175}{50}$ = 3.5

∴ Dionne can fold 210 packing boxes = $\dfrac{210}{3.5}$ minutes

= 60 minutes

Also,

Elias can fold 120 packing boxes = 40 minutes

∴ In 1 mìnute, Elias can fold number of packing boxes = $\dfrac{120}{40}$ = 3

∴ Elias can fold 210 packing boxes = $\dfrac{210}{3}$ minutes

= 70 minutes

Thus, Dionne will fold 210 packing boxes in "60 mìnutes" and

Elias will fold 210 packing boxes in "70 mìnutes".

8 0

3 years ago

Line segment ON is perpendicular to line segment ML.

3241004551 [841]

Answer:

2 units.

Step-by-step explanation:

I just took the Unit Test.

6 0

3 years ago

Read 2 more answers

BRAINLIEST ASAP! PLEASE HELP ME :)

jenyasd209 [6]

<h3>Answers:</h3><h3>There are four solutions and they are</h3>

$\theta = 0$
$\theta = \pi$
$\theta = \frac{7\pi}{6}$ ... this says "7pi over 6"
$\theta = \frac{11\pi}{6}$ ... this says "11pi over 6"

===========================================

Work Shown:

$\sin(\theta)+1 = \cos(2\theta)\\\\\sin(\theta)+1 = 1-2\sin^2(\theta)\\\\\sin(\theta)+1-1+2\sin^2(\theta)=0\\\\2\sin^2(\theta)+\sin(\theta)=0\\\\\sin(\theta)(2\sin(\theta)+1)=0\\\\\sin(\theta)=0 \ \text{ or } \ 2\sin(\theta)+1=0\\\\$

----------

Solving $\sin(\theta)=0$ leads to $\theta=n\pi$ for any integer n.

----------

Solving $2\sin(\theta)+1=0$ leads to...

$2\sin(\theta)+1=0\\\\2\sin(\theta)=-1\\\\\sin(\theta)=-\frac{1}{2}\\\\\theta=\arcsin\left(-\frac{1}{2}\right) \ \text{ or } \ \theta=\pi-\arcsin\left(-\frac{1}{2}\right)\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\pi-\frac{11}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{-5}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{7}{6}\pi+2\pi n\\\\$

-----------

The general solution set is

$\{\theta=n*\pi, \theta=\frac{11}{6}\pi+2\pi n, \theta=\frac{7}{6}\pi+2\pi n\}$

Again, n is any integer.

Let's look at a table of values where we plug in various integers for n. See the attached image below. Note the stuff in the highlighted yellow cells represents expressions that are between 0 and 2pi = 6.28

Therefore the four solutions are $\{0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\}$ after we plug the proper values of n into the expressions, to have things match what the table shows.

6 0

3 years ago

What’s the reason for the fourth statement in this proof

Ksivusya [100]

What statement.........

8 0

3 years ago