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RoseWind [281]
3 years ago
14

Please answer 2) multiple choice questions correctly I need it today please show work .

Mathematics
1 answer:
MrRissso [65]3 years ago
5 0

Answer:

Step-by-step explanation:

20. First find the values of the numbers being added.

(\frac{3}{4})^3 = \frac{27}{64}

(\frac{5}{8} )^2 = \frac{25}{64}

These numbers have a common denominator, so you can add the numerators

27 + 25 = 52

so the number comes out to \frac{52}{64}

to simplify, divide the numerator and denominator by 4.

So the final fraction becomes \frac{13}{16}

However the question doesn't have this answer and now I am confused. I assumed that the expression was r^{3}+s^{2} but the picture is a little bit blurry and I have bad eyesight so let me know if it is not. I'm too tired to delete this so next problem.

21.

4x + 8x + 7 + 4

Combine the like terms to get the expression

12x + 11 the answer is c.

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Step-by-step explanation:

Point A represents the integer 5.

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Given sintheta =7/11 and sectheta is less than 0, find costheta and tantheta
Korvikt [17]

Answer:

Step-by-step explanation:

Sin theta is the ratio of side opposite over hypotenuse of a reference angle situated at the origin in an x-y coordinate plane.  If sec theta is negative, then the only quadrant where sin is positive AND sec is negative is quadrant 2.  Remember that sec theta is the inverse of cos theta.  Puttling our right triangle in QII, the side measuring 7 is across from the angle and the hypotenuse is 11.  In order to find the cos theta and tan theta, we need the side adjacent to the angle.  Use Pythagorean's Theorem to find the side adjacent.

11^2=7^2+x^2 and

121=47+x^2 and

72=x^2 so

x=6\sqrt{2}

Remember that this value is why the sec is negative.  Because x is negative in QII, the cos theta is side adjacent over hypotenuse:

cos\theta=-\frac{6\sqrt{2} }{11}  and

tan\theta=-\frac{7}{6\sqrt{2} }

But we should probably rationalize that denominator, so

tan\theta=-\frac{7}{6\sqrt{2} }*\frac{\sqrt{2} }{\sqrt{2} } =-\frac{7\sqrt{2} }{12}

5 0
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