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tester [92]
3 years ago
7

Statement Reason

Mathematics
1 answer:
masya89 [10]3 years ago
4 0

Answer:

D for Plato

Step-by-step explanation:

D. Corresponding side lengths in similar triangles are proportional in length

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lana66690 [7]
The answer would be $7050.00.
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2 years ago
Question 1(Multiple Choice Worth 5 points)
Nadusha1986 [10]

Answer: arc\ length =13.19cm

Step-by-step explanation:

You need to use the following formula:

arc\ length=r\theta

Where "r" is the radius and \theta is is the central angle of the arc in radians.

Assuming that that the angle \frac{3\pi }{5} is the central angle of the arc, and knowing that the radius is 7 centimeters, you can substitute values into the formula.

Therefore, you get:

arc\ length=r\theta\\\\arc\ length= (7cm)(\frac{3\pi}{5})}\\\\arc\ length =13.19cm

4 0
3 years ago
Who procrastinates to study✋
IrinaK [193]
Hi There! :)

<span>Who procrastinates to study

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5 0
3 years ago
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}}

4 0
3 years ago
What is the product of the polynomials below? (6x^2-3x-6)(4x^2+5x+4)​
Lunna [17]

Answer:

(6 - 5x(2))(x(4) - x(3))

(6 - 5x^2) (x^4 - x^3)

6x^4 - 6x^3 - 5x^6 + 5x^5

-5x^6 + 5x^5 + 6x^4 - 6x^3

Step-by-step explanation:

4 0
3 years ago
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