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2 years ago

Work out the area of the shaded shape.

Mathematics

1 answer:

Step2247 [10]2 years ago

4 0

Answer:

Step-by-step explanation:

A = 8(14) - 9(8 - 4) = 76 m²

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Complete each sentence 4% of __days is 56 days

iris [78.8K]

1400 days = 56 days... 100 over 4 * 4/100*d=100 over 4 * 56 xd= 5600 over 4 d = 1400

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2 years ago

N to the second power - 5 in - 1 in equals 6

Wittaler [7]

Answer:

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3 years ago

Can anyone help me with this question

Lina20 [59]

Answer:

$78$

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2 years ago

In ΔVWX, the measure of ∠X=90°, XW = 65, WV = 97, and VX = 72. What is the value of the cosine of ∠V to the nearest hundredth?

bonufazy [111]

Answer:

The cosine of ∠V is of 0.74.

Step-by-step explanation:

Relations in a right triangle:

The cosine of an angle is given by the length of the adjacent side divided by the length of the hypotenuse.

XW = 65, WV = 97, and VX = 72.

$\sqrt{65^2+72^2} = 97$ , and thus, this is a right triangle.

What is the value of the cosine of ∠V to the nearest hundredth?

The hypotenuse is the largest side, that is, WV = 97.

The adjacent side of angle V is VX = 72. So

$\cos{B} = \frac{72}{97} = 0.74$

The cosine of ∠V is of 0.74.

8 0

3 years ago

Find the average rate of change of the function over the given interval

sattari [20]

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\$

$\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\$

$\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }$

8 0

3 years ago