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Natalija [7]
4 years ago
15

Solve the problem.

Mathematics
1 answer:
Lorico [155]4 years ago
5 0

Answer:

<h2>560</h2>

Step-by-step explanation:

You must use a combination:

_nC_k=\dfrac{n!}{k!(n-k)!}

We have <em>n = 16</em>, <em>k = 3</em>.

Substitute:

_{16}C_3=\dfrac{16!}{3!(16-3)!}=\dfrac{13!\cdot14\cdot15\cdot16}{2\cdot3\cdot13!}\qquad\text{cancel}\ 13!\\\\=\dfrac{14\cdot15\cdot16}{2\cdot3}=\dfrac{7\cdot5\cdot16}{1}=560

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Find x+y+z from the following figure<br><br><br>​
hoa [83]

Step-by-step explanation:

the inner top vertex angle seems to be 90°, so x = 90° too. all angles around a single point in one side of a line are in total 180°.

also, remember, all angles inside a triangle always sum up to 180°.

so, the third inner angle of the triangle is

180 - 90 - 40 = 50°.

that makes z = 180 - 50 = 130°

and y = 180 - 40 = 140°.

8 0
2 years ago
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grin007 [14]

Answer:

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3 years ago
Divide 750 into the ratio 2:3​
Anestetic [448]

Answer:

Answer:22:37

DECIMAL: 10.38

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Step-by-step explanation:

HOPE IT HELPS!!!!!

3 0
3 years ago
Read 2 more answers
For ΔABC, ∠A = 8x - 10, ∠B = 10x - 40, and ∠C = 3x + 20. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C'
Sholpan [36]

Answer: It does not confirm that ΔABC∼ΔA'B'C by the AA criterion as the measure of angle B is negative which is not possible.

Step-by-step explanation:

Since we have given that

In ΔABC,

∠A = 8x - 10, ∠B = 10x - 40, and ∠C = 3x + 20

In ΔA'B'C',

∠A' = 6x + 10, ∠B' = 70 - x, and ∠C' = 10x 2

Since ΔABC gets a dilation by a scale factor of \dfrac{1}{2}

So, it becomes,

\dfrac{\angle A}{\angle A'}=\dfrac{1}{2}\\\\\dfrac{8x-10}{6x+10}=\dfrac{1}{2}\\\\2(8x-10)=6x+10\\\\16x-20=6x+10\\\\16x-6x=10+20\\\\10x=30\\\\x=\dfrac{30}{10}=3

Now, put the value of

\angle B=10(3)-40=30-40=-10\\\\and \angle B'=70-30=40

It does not confirm that ΔABC∼ΔA'B'C by the AA criterion as the measure of angle B is negative which is not possible.

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3 years ago
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Sindrei [870]

Answer:

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Step-by-step explanation:

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3 years ago
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