Which describes the effect of the transformations on the graph of ƒ(x) = x2 when changed to ƒ(x) = x^2+2

Finn bought 2 packs of stickers. A friend gave him 4 more stickers. Now he has 24 stickers in all. How many stickers were in eac

Murljashka [212]

2x+4=24 -4 -4 ------------- 2x=20 ÷2 ÷2 ---------- X=10 So 10 stickers in each pack

6 0

3 years ago

Read 2 more answers

Help me solve this question please

Vanyuwa [196]

Answer:

Step-by-step explanation:

3 0

2 years ago

NEED HELP

KIM [24]

A coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).

<h3>What is directly proportional?</h3>

It is defined as the relationship between two quantities as one quantity increases the other quantity also increases and vice versa.

It is given that there is a proportional relationship between cost and pints of blueberries as,

Blueberries (in pints) 2 5

Cost (in dollars) 6.80 17

17/6.80 = 5/2

As we draw the graph and analyze each option one by one we get the graph shown by the statement C following the given proportional relationship.

Thus, a coordinate plane with the x-axis labeled Blueberries (in pints) and the y-axis labeled Cost (in dollars) shows a line going through (0, 0) and (5, 17).

Learn more about the directly proportional here:

brainly.com/question/16937826

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4 0

1 year ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so