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3 years ago

Can someone answer #13, #15, and #16

Mathematics

2 answers:

Dominik [7]3 years ago

8 0

#13: b
#14: c
#16: c

It is very easy, have a nice day :)

Mars2501 [29]3 years ago

3 0

16). Volume of a cube: v = s³ (s) stands for side length

v = 3³
v = 3 × 3
v = 9

You were right, the answer is a.
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I drew a picture to figure out this problem. I drew a circle with 16 different sections. Half of them were black and the other half were white. If a dart is thrown 3 times and lands on the black each time, then you have to eliminate the 3 times. So on the picture, eliminate 3 black blocks. The question wants to know the probability of the dart landing on a black if it's thrown a fourth time. In order to find this out, you need to eliminate just one more black section. Then, you have four more black sections left. You have to put 4 over the number of total black sections there were, which are 8. 4/8 = 1/2, so the final answer is 50%. Sorry if this is confusing.

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Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$