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3 years ago

What part of the circle is shaded?

Mathematics

1 answer:

IceJOKER [234]3 years ago

3 0

Answer:

part a if that's what you mean

Step-by-step explanation:

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Jack was 8 years older than pricillla. Together their ages totaled 166. what are their ages

mylen [45]

The answer is 58 I think 166-8= 158

4 0

3 years ago

Read 2 more answers

What is the solution to:<br><img src="https://tex.z-dn.net/?f=6%20%5Cgeqslant%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20%2B%2021" id="Tex

Vladimir79 [104]

I really hope this helps. Btw x<-30

7 0

3 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago

For the month of January in a certain city, 42% of the days are cloudy. Also in the month of January in the same city, 41% o

zalisa [80]

Answer: 0.9762

Step-by-step explanation:

Let A be the event that days are cloudy and B be the event that days are rainy for January month .

Given : The probability that the days are cloudy = $P(A)=0.42$

The probability that the days are cloudy and rainy = $P(A\cap B)=0.41$

Now, the conditional probability that a randomly selected day in January will be rainy if it is cloudy is given by :-

$P(B|A)=\dfrac{P(B\cap A)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{0.41}{0.42}=0.97619047619\approx0.9762$

Hence, the probability that a randomly selected day in January will be rainy if it is cloudy = 0.9762

7 0

4 years ago

5x5x5x5x5x5x5x5x10x10x00.0001

gulaghasi [49]

3906.25

Use a calculator

4 0

3 years ago