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Nina [5.8K]
2 years ago
11

HELP ASAP. (please show work) solve for x.

Mathematics
1 answer:
Fantom [35]2 years ago
4 0

Answer:

x = 11

Step-by-step explanation:

a straight line = 180 degrees . 180-125=55 . 55/5 = 11

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The radius of one circle is three the time radius of another circle. The sum of their areas is 12pi. Find the radius of each cir
zlopas [31]
   
\displaystyle\\
\begin{cases}
R_1 = 3R_2\\
\pi R_1^2 +\pi R_2^2=12\pi ~~~\Big|~~:\pi 
\end{cases} \\  \\ 
\begin{cases} 
R_1 = 3R_2\\
R_1^2 + R_2^2=12 
\end{cases} \\  \\ 
(3R_2)^2+ R_2^2=12 \\  \\ 
9R_2^2 + R_2^2=12 \\  \\ 
10R_2^2=12\\\\
R_2^2 = \frac{12}{10} = \frac{6}{5}  \\  \\ 
R_2 =  \boxed{\sqrt{\frac{6}{5}} } \\  \\ 
R_1 = \boxed{3\sqrt{\frac{6}{5}} }



7 0
3 years ago
Identify the function that contains the data in the following table:Х-20o235f(x)531224O f(x) = x/ + 1O f(x) = x - 21O f(x) = lx
kirill [66]

To identify the function that contains the data in the table, we should first visualize the data.

A graph of the data is shown below:

From the plot above, we can identify that:

The graph above is a graph of f(x) = |x|, translated to the right 2 units and translated upwards 1 unit.

Hence, the function is:

f(x)\text{  = |x-2| + 1}

Answer:

Option D

3 0
1 year ago
Trigonometry help!! - double angle formulae
ivolga24 [154]

Answer:

The two rules we need to use are:

Sin(a + b) = sin(a)*cos(b) + sin(b)*cos(a)

cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)

And we also know that:

sin^2(a) + cos^2(a) = 1

To solve the relations, we start with the left side and try to construct the right side.

a) Sin(3*A) = sin (2*A + A) = sin(2*A)*cos(A) + sin(A)*cos(2*A)

sin(A + A)*cos(A) + sin(A)*cos(A + A)

(sin(A)*cos(A) + sin(A)*cos(A))*cos(A) + sin(A)*(cos(A)*cos(A) - sin(A)*sin(A))

sin(A)*cos^2(A) + sin(A)*cos^2(A) + sin(A)*cos^2(A) - sin^3(A)

3*sin(A)*cos^2(A) - sin(A)*sin^2(A)

sin(A)*(3*cos^2(A) - sin^2(A))

Now we can add and subtract 4*sin^3(A)

sin(A)*(3*cos^2(A) - sin^2(A)) + 4*sin^3(A) -  4*sin^3(A)

sin(A)*(3*cos^2(A) + 3*sin^2(A)) - 4*sin^3(A)

sin(A)*3*(cos^2(A) + sin^2(A)) - 4*sin^3(A)

3*sin(A) - 4*sin^3(A)

b) Here we do the same as before:

cos(3*A) = 4*cos^3(A) - 3*cos(A)

We start with:

Cos(2*A + A) =  cos(2*A)*cos(A) - sin(2*A)*sin(A)

= cos(A + A)*cos(A) - sin(A + A)*sin(A)

= (cos(A)*cos(A) - sin(A)*sin(A))*cos(A) - ( sin(A)*cos(A) + sin(A)*cos(A))*sin(A)

= (cos^2(A) - sin^2(A))*cos(A) - sin^2(A)*cos(A) - sin^2(A)*cos(A)

= cos^3(A) - 3*sin^2(A)*cos(A)

=  cos(A)*(cos^2(A) - 3*sin^2(A))

now we subtract and add 4*cos^3(A)

= cos(A)*(cos^2(A) - 3*sin^2(A)) + 4*cos^3(A) - 4*cos^3(A)

= cos(A)*(-3*cos^2(A) - 3*sin^2(A)) + 4*cos^3(A)

= cos(A)*(-3)*(cos^2(A) + sin^2(A)) + 4*cos^3(A)

= -3*cos(A) + 4*cos^3(A)

8 0
3 years ago
Granola
Semenov [28]

Answer:

1/8 : 1/2

Step-by-step explanation:

Is that the right answer? I’m not sure lol I don’t remember

4 0
3 years ago
If sin theta=-square root 3 over 2 and pie
choli [55]

Answer:cos theta = -(1)/2; tan theta = sqrt(3)

cos theta = -(1)/2; tan theta = -1

cos theta = (sqrt(3))/4; tan theta = -2

cos theta = 1/2; tan theta = sqrt(3)

cos theta = 4/3

Step-by-step explanation:

7 0
3 years ago
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