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lord [1]
2 years ago
15

My girl friends face whats 7x7

Mathematics
2 answers:
jarptica [38.1K]2 years ago
6 0

Answer: this is 2 grade stuff but ok... 49

weqwewe [10]2 years ago
3 0

Answer: Why is the basic math question is here? But the answer is 49, don’t put the basic expression next time!

Step-by-step explanation:

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How do I complete these questions a bit confused . (extra credit work )​
otez555 [7]

Answer:

(1)

a = \frac{3\sqrt 3}{2}

b = \frac{3}{2}

(2)

a = \sqrt 6

b = \sqrt 2

Step-by-step explanation:

Solving (1):

Considering

\theta = 60^o

We have:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

This gives:

\sin(60^o) = \frac{a}{3}

Solve for a

a = 3 * \sin(60^o)

\sin(60^o) = \frac{\sqrt 3}{2}

So:

a = 3 * \frac{\sqrt 3}{2}

a = \frac{3\sqrt 3}{2}

To solve for b, we make use of Pythagoras theorem

3^2 = a^2 + b^2

This gives

3^2 = (\frac{3\sqrt 3}{2})^2 + b^2

9 = \frac{9*3}{4} + b^2

9 = \frac{27}{4} + b^2

Collect like terms

b^2 = 9 - \frac{27}{4}

Take LCM and solve

b^2 = \frac{36 - 27}{4}

b^2 = \frac{9}{4}

Take square roots

b = \frac{3}{2}

Solving (2):

Considering

\theta = 60^o

We have:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

This gives:

\sin(60^o) = \frac{a}{2\sqrt 2}

Solve for a

a = 2\sqrt 2 * \sin(60^o)

\sin(60^o) = \frac{\sqrt 3}{2}

So:

a = 2\sqrt 2 * \frac{\sqrt 3}{2}

a = \sqrt 2 * \sqrt 3

a = \sqrt 6

To solve for b, we make use of Pythagoras theorem

(2\sqrt 2)^2 = a^2 + b^2

This gives

(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2

8 = 6 + b^2

Collect like terms

b^2 = 8 - 6

b^2 = 2

Take square roots

b = \sqrt 2

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3 years ago
Can you factorise<br> X=3a-2x10-2a
tamaranim1 [39]

x = 3a - 2·10 - 2a

x = 3a - 2a - 20

x = a - 20

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