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3 years ago

5

Help me plzzz!?!?? true or false

2 answers:

Anon25 [30]3 years ago

5 0

A. False - She has 15 plants

B. True - She has 4 plants that are 7-9 and 4 plants that are 10-12

C. True - Medain means middle, and since it is an even number, and it's between 9 and 10, there's a good chance it's around 11

Mrac [35]3 years ago

3 0

Answer:

false, true, false

Step-by-step explanation:

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A bag contains 1 purple beads and 3 green beads. A bead is drawn and then replaced before drawing the second bead. Find the prob

disa [49]

ggg

Step-by-step explanation:

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3 years ago

Read 2 more answers

It costs you $22.25 per day to rent a car in New York City. In addition, you have to pay 16 cents for every mile you travel. How

olganol [36]

Answer: $178.13

Step-by-step explanation:

418x0.16=66.88

22.25x5=111.25

111.25+66.88=178.13

4 0

3 years ago

Help will mark brainlest

3241004551 [841]

Answer:

83

Step-by-step explanation:

82+87+83+77+70+80+88= 567

Then you divide that answer by the number of numbers you added by which is 7.

567/7= 81 so 83 was the missing number

6 0

3 years ago

Consider the polar curve r = 4sin(2θ). What is the length of each petal, and how many petals does the graph have? a The length o

VashaNatasha [74]

Answer:

d The length of each petal is 4, and the number of petals is 4.

Step-by-step explanation:

3 0

2 years 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

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago