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3 years ago

How do I find the answer to m

Mathematics

1 answer:

ZanzabumX [31]3 years ago

3 0

Those are vertical angles,so they are equal
150=5m
m=150/5
m=30

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16y + 0 = 16y

balu736 [363]

The answer is C. Identify property of addition

4 0

3 years ago

4.3 Mr Mokwebo 's two sons, Refentse and Tokelo have just passed grade 7 and grade 9 respectively. He will give them some money

Bumek [7]

The amount of money each of Mr Mokwebo 's two sons; Tokelo and Refentse will get is R116 and R95 respectively.

<h3>How to write and solve equation:</h3>

Amount needed to buy chocolate = R253
Tokelo's money = x
Refentse's money = x - 21

Total amount needed = Tokelo's money + Refentse's money

253 = x + (x - 21)

253 = x + x - 21

253 = 2x - 21

253 - 21 = 2x

232 = 2x

divide both sides by 2

x = 232/2

x = 116

Therefore,

Tokelo's money = x = R116

Refentse's money = x - 21

= 116 - 21

= R95

Learn more about equation:

brainly.com/question/2972832

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6 0

2 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago

Pls answer this asap

Serga [27]

Answer:

its c or d

Step-by-step explanation:

6 0

3 years ago

The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If

frutty [35]

For this case we have the following equation:
$w = 0.9* \sqrt{E}$
Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:
$w = 0.9* \sqrt{12}$
Making the corresponding calculation:
$w=3.12$
Answer:
the approximate weight of the spring in pounds is:
$w=3.12$

4 0

3 years ago

Read 2 more answers