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Crank
3 years ago
15

A group of students is collecting books to add to their librarby. The table shows the number of books in the librarby after 1, 3

, and 5 days. If the relationship between days and books continues to be linear, which ordered pairs could appear in the table? Select all that apply. (You like my Undertale reference? Will give Brainliest if you provide an explanation)
Here are the points it gives: (0,8), (2,23), (4,32), (6,48), and (7,48)

Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
8 0
I got 7,48 that’s correct
Grace [21]3 years ago
6 0

Answer:

(7,48)

Step-by-step explanation

the books go up by ten  and the days go up by 2

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Answer:

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3 years ago
Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))
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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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Step-by-step explanation:

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