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ycow [4]
3 years ago
10

Solve problem in picture​

Mathematics
1 answer:
dsp733 years ago
6 0

Answer:

this image is not clearly visible try another

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Use the GCF to factor 8-6
soldier1979 [14.2K]
The GCF of 8 and 6 is two
4 0
3 years ago
Can someone solve this and explain?
Anarel [89]
Answer: (y-c)/m
Reasoning:
Subtract c from both sides [ y-c=mx+c-c ]
Divide m from both sides [ (y-c)/m=mx/m ]
7 0
3 years ago
Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
3 years ago
What is the value of side "a" in the triangle shown below?
maks197457 [2]

Answer:

a = 5

Step-by-step explanation:

Since the triangle is right use the sine ratio to solve for a and

sin30° = \frac{1}{2}, thus

sin30° = \frac{opposite}{hypotenuse} = \frac{a}{10}, hence

\frac{a}{10} = \frac{1}{2} ( cross- multiply )

2a = 10 ( divide both sides by 2 )

a = 5

6 0
3 years ago
What is the midpoint between A(-6,1) and B(0,2)?
Elan Coil [88]

Answer:

(-3, 3/2)

Step-by-step explanation:

To find the midpoint between two points you are going to add the X1 and X2, then divide by two. Then you are going to add the Y1 and Y2, and divide by two.

A(-6,1)       B(0,2)

= (-6+0, 1+2)

= (-6, 3)

=(-6/2, 3/2)

=(-3, 3/2)

7 0
3 years ago
Read 2 more answers
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