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3 years ago

13

Kate is saving to take a SAT prep course that costs $350. So far, she has saved $180, and she adds &17 to her savings each w

eek. How many more weeks just she save to be able to afford the course?

1 answer:

Thepotemich [5.8K]3 years ago

4 0

The Answer

10 Weeks

Step-by-step explanation:

180+170 = 350

17x10 = 170

Would take 10 weeks of saving 17 dollars a week to become equal to 350.

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Answer:

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3 years ago

If r=10 and s=31, find R. Round to the nearest tenth.

mars1129 [50]

Answer:

C. R = 17.9 degrees

Step-by-step explanation:

We have a rectangle triangle with the adjacent side and the opposite side (neither of which are the hypotenuse).

The relation between those elements is the tangent:

$tan(angle) = \frac{Opposite side}{Adjacent side}$

So, to isolate the angle, we modify the formula as such:

$angle = arctan(\frac{Opposite side}{Adjacent side}) = arctan(\frac{10}{31}) = arctan(0.3225) = 17.87$

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3 years ago

Read 2 more answers

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

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sertanlavr [38]

Answer:

$x=4\\\\y=6\\\\IK=47$

Step-by-step explanation:

Two Tangent Theorem: Tangents which meet at same point are equal in length.

Here tangents at J and H meet at I.

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