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babunello [35]
3 years ago
12

PLS HELP Use the divisibility rules to determine which of the following numbers 315 is divisible by.(Check all that apply.) 2 3

5 7 11
Mathematics
1 answer:
Vlad [161]3 years ago
6 0

we can check divisibility of 315

Divisibility by 2:

We can see that last digit of 315 is 5

and it is not multiple of 2

so, it is not divisble by 2

Divisibility by 3:

We will all digits

3+1+5=9

and we know that 9 is divisible by 3

so, 315 is divisible by 3

Divisibility by 5:

We can see that last digit of 315 is 5

and it is  multiple of 5

so, it is  divisble by 5

Divisibility by 7:

take frst two digit without last digit

is 31

subtract last digit times 2

2*5=10

difference is 31-10=21

since, 21 is divisible 7

so, 315 is divisible by 7

Divisibility by 11:

digits at odd numbers are 1

odd=1

digits at even numbers are 5,3

even=5+3=8

difference=even-odd

difference=8-1=7

since, 7 is not multiple of 11

so, 315 is not divisible by 11



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First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

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Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

\dfrac{b}{\sin B} = \dfrac{c}{\sin C}

\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}

\sin C = \dfrac{8.9\sin 63^\circ}{10}

C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}

C \approx 52.5^\circ

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.

Answer: one triangle

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Answer:

The answer is C

Step-by-step explanation:

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3 years ago
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