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saw5 [17]
3 years ago
5

Answer all questions please

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
8 0
A) Base Angle
B) Leg
C) Vertex
D) Leg
E) Base Angle
F) Base

5. SSS because there are no angles and the three lines would go through the middle line. There is no SSA.
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Solve the equation: sin(2x)= sin x
Anna [14]

Answer:

Below

Step-by-step explanation:

sin(2x) = 2 ×cos(x)× sin(x)

● sin(x) = 2 × cos(x) × sin(x)

● 2 × cos(x) = 1

● cos (x) = 1/2

So we can deduce that:

● x = Pi/3 + 2*k*Pi

● or x = -Pi/3 + 2*k*Pi

K is an integer

3 0
3 years ago
M/5 + 0.12 = -6.3 (How do you do this?)
deff fn [24]

Answer:

m = -32.10

Step-by-step explanation:

You solve this the same way you do any two-step linear equation.

1. Subtract the constant that is on the side of the equal sign with the variable term:

m/5 = -6.42

2. Multiply by the inverse of the coefficient of the variable. That means multiply by 5 in this case.

m = 5·(-6.42) = -32.10

_____

Of course, the rules of equality tell you whatever you do to one side of the equation must also be done to the other side. So, when we say subtract 0.12, we mean subtract 0.12 from both sides of the equation. The same for multiplication or division.

4 0
4 years ago
Read 2 more answers
(08.01)Consider the following system of equations:
amid [387]
The first one is correct.
3 0
4 years ago
Read 2 more answers
VW is the bisector of AY , and they intersect at E. If EY = 3.5, what is AY?
frosja888 [35]

Step-by-step explanation:

this is solutions of your questions

4 0
3 years ago
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

3 0
3 years ago
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