Ask question

Ask question

All categories

2 years ago

10

please answer fast!!!! The wheels on Jessica's bike are 64 inches in circumference. How many times do the wheels rotate if Jessi

ca rides 300 yards?

1 answer:

MAXImum [283]2 years ago

8 0

Answer:

169

Step-by-step explanation:

You might be interested in

The product of -4 and the number is greater than or equal to 8

pishuonlain [190]

Answer:

it would be -2 or higher

4 0

3 years ago

What could be the answer to (n+4)+4

Sav [38]

Answer:

=N+8

Step-by-step explanation:

You would remove the parenthesis

Add the numbers together

and the n would = 8

8 0

3 years ago

Please help im offering the highest amount of points I can give.

STatiana [176]

Answer:

96 cm^3

Step-by-step explanation:

I hope this helped!

5 0

3 years ago

Given: 2x + 11 = 15 Prove: x = 2 Statements Reason 1. 2x + 11 = 15 1. Given 2. 2x = 4 2. 3. X = 2 3. Division Property of Equali

wariber [46]

Answer:

Statements Reasons

1. 2x + 11 = 15 1. Given

2. 2x = 4 2. Subtraction Property of Equality

3. X = 2 3. Division Property of Equality

Step-by-step explanation:

An equation can be solved and its solution proven using algebraic theorems and properties. To create a proof, form two columns. Label one side Statements and the other Reasons.

Begin your proof listing the any information given to you. List as the reason - Given.

Then list the next step which here would be to subtract by 11 on both side. The reason is Subtraction Property of Equality. Subtraction is the inverse of addition. Inverse axiom is another acceptable reason.

Then divide both sides by 2. The reason is Division Property of Equality or Inverse axiom once again. See the proof below.

Statements Reasons

1. 2x + 11 = 15 1. Given

2. 2x = 4 2. Subtraction Property of Equality

3. X = 2 3. Division Property of Equality

6 0

3 years ago

Read 2 more answers

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