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Mariana [72]
3 years ago
7

Which of the following numbers are not divisible by 2,3,4,5,6,9 or 10?

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0
What are the numbers?
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B-c,when b=8 and c=7
elixir [45]
If I understand your question correctly then it's simple all you have to do is use the numbers in the place of the b and c so the question would be what is 8-7 which equals 1 
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3 years ago
HELP AND PLZZZ EXPLAIN!
LuckyWell [14K]

Simplify 6 - 25 to -19

18 + 3| -19 | - 11

Simplify | -19 | to 19

18 + 3 × 19 - 11

Simplify 3 × 19 to 57

18 + 57 - 11

Simplify

<u>= 64</u>

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3 years ago
Which is a qualitative graph?
larisa86 [58]

Answer: Qualitative graphs are graphs that are used to represent situations that do not necessarily have numerical values. Qualitative graphs represent the essential elements of a situation in a graphical form. For example, Graph A could represent a car that is accelerating at a constant rate.

3 0
4 years ago
Read 2 more answers
Liz and Mike started exercising at 4:15 p.m. Mike exercised for 90 minutes. Liz exercised half an hour less than Mike.
Sever21 [200]

Based on the information given, it should be noted that Liz will stop exercising at 5.15 p.m.

Since Mike exercised for 90 minutes and Liz exercised half an hour less than Mike. Then, the number of minutes used by Liz will be:

= 90 - (1/2 × 60)

= 90 - 30

= 60 minutes = 1 hour

Now, the time that Liz stopped exercising will be

= 4.15pm + 1 hour

= 5.15 pm

Liz will stop exercising at 5.15 p.m.

Read related link on:

brainly.com/question/25337748

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%202%20%20%5Csin%28%20%5Calpha%20%29%20-%20%20%20%5Ccos%28%20%5Calpha%20%29%20%20%3D%202%20%5C
mr Goodwill [35]

2\sin\alpha-\cos\alpha=2

Consider the substitution \tan\dfrac\alpha2=\beta. Then by the double angle identities we get

\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2

\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2

We also have

\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}

so that

\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}

\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}

and the original equation has been transformed to

\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2

Solve for \beta:

5\beta^2-1=2+2\beta^2

3\beta^2=3

\beta^2=1

\beta=\pm1

Solving for \alpha gives

\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi

\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi

where n is any integer. Both \sin and \cos are 2\pi-periodic, which is to say

\cos(x+2n\pi)=\cos x

\sin(x+2n\pi)=\sin x

so that

\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1

\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0

and we find that

\sin\alpha+2\cos\alpha=\pm1

3 0
4 years ago
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