PLEASE HELP!!!!!!!!!

1 answer:

7 0

-- 100% of the original number is ALL of it.

-- If it's increased by that amount, it becomes DOUBLE
(twice as much as) what it was originally.

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How to find inequalities

geniusboy [140]

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6 0

3 years ago

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Nostrana [21]

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3 0

3 years ago

-129 = -48 - n <br> Thank you

Flauer [41]

−129=−48−n

Simplify:

−129=−48+−n

Flip the equation.

−n−48=−129

Add 48 to both sides.

−n−48+48=−129+48

−n=−81

Divide both sides by -1.

−n/−1 = −81/−1

n=81

3 0

3 years ago

PLEASE HELP WITH SIGNIFICANT FIGURES *AWARDING BRAINEST ANSWER* ASAP

marishachu [46]

1) 0.07 zeros only matter if there is a non-zero digit, as well as the decimal to the left of it
2)90,000 it would have been 89000 if the 972 had been an 872
3).00420 must include the last zero, it is considered significant because of the reason in number one

4 0

3 years ago

How do you solve cos(pi/24) using Half-Angle formulas, and leaving in simplified form?

adoni [48]

$\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}
\\
\\ \cos{\frac{\pi}{24}}=\cos{\frac{\frac{\pi}{12}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{12}}{2}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2} } 
\\
\\ \cos{\frac{\pi}{12}}=\cos{\frac{\frac{\pi}{6}}{2}}=\sqrt{\frac{1+\cos \frac{\pi}{6}}{2}}= \sqrt{ \frac{1}{2}+ \frac{ \frac{ \sqrt{3} }{2} }{2} } =\sqrt{ \frac{2}{4}+ \frac{ \sqrt{3} }{4} } = \sqrt{\frac{ 2+\sqrt{3} }{4} } = 
\\
\\ =\frac{ \sqrt{2+\sqrt{3}} }{2} 
$

$\cos{\frac{\pi}{24}}= \sqrt{ \frac{1}{2}+ \frac{\cos \frac{\pi}{12}}{2} } = \sqrt{ \frac{1}{2}+ \frac{\frac{ \sqrt{2+\sqrt{3}} }{2}}{2} } = \sqrt{ \frac{2}{4}+ \frac{ \sqrt{2+\sqrt{3}} }{4}} } =\sqrt{ \frac{ 2+\sqrt{2+\sqrt{3}} }{4}} } 
\\
\\\cos{\frac{\pi}{24}}=\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}} {2}$

6 0

3 years ago