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frozen [14]
2 years ago
9

What is the radius of this circle?

Mathematics
2 answers:
ahrayia [7]2 years ago
5 0

Answer:

10

Step-by-step explanation:

Vesna [10]2 years ago
3 0

Answer:

C

Step-by-step explanation:

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1 the sum of the binomials 3 + 2x and 4 – 5x. Explain how you can use the
GalinKa [24]

Answer:

(3 + 2x) + (4 – 5x) = 7 - 3x

(3 + 2i) + (4 – 5i) = 7 - 3i

Step-by-step explanation:

In the addition of two binomials you have to add separately each monomial as follows:

(3 + 2x) + (4 – 5x) =

= (3 + 4) + (2x - 5x) =

= 7 - 3x

Analogously, the sum of the complex numbers 3 + 2i and 4 – 5i is made adding the two real parts from one side, and the two imaginary parts from the other side, as follows:

(3 + 2i) + (4 – 5i) =

= (3 + 4) + (2i - 5i) =

= 7 - 3i

7 0
3 years ago
Read 2 more answers
Solve for the following system by using Elimination4x-2y=86x+6y=30
Ilia_Sergeevich [38]

System

\begin{gathered} 4x-2y=8 \\ 6x+6y=30 \end{gathered}\begin{gathered} 3(4x-2y)=8 \\ 6x+6y=30 \\  \\ 12x-6y=24 \\ + \\ 6x+6y=30 \\  \\ 12x+6x-6y+6y=24+30 \\ 18x=54 \\ x=3 \end{gathered}

Now, for y

\begin{gathered} 4x-2y=8 \\ -2y=8-4x \\ 2y=4x-8 \\ y=\frac{4x-8}{2} \\ y=\frac{4\cdot3-8}{2} \\ y=\frac{12-8}{2} \\ y=\frac{4}{2} \\ y=2 \end{gathered}

3 0
1 year ago
Solve: (a-7)(a+1) = 0
Svetradugi [14.3K]

Answer:

0

Step-by-step explanation:

7 0
3 years ago
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Find the coordinates of point T
KATRIN_1 [288]

Answer:

(2,0)

Step-by-step explanation:

(6-(10-6), 4- (8-4))

8 0
3 years ago
Write two expressions with unlike denominators whose sum is x-3/x+2, I need help it is confusing for me.
skad [1K]

Answer:

\displaystyle A=\frac{x-2}{x+3}

\displaystyle B=\frac{-5}{(x+3)(x+2)}

Step-by-step explanation:

We need to find two expressions with unlike denominators what sum

\displaystyle S=\frac{x-3}{x+2}

Let's suppose one of the expressions is:

\displaystyle A=\frac{x-2}{x+3}

Now we subtract S minus A to find the other expression B:

\displaystyle B=S-A=\frac{x-3}{x+2}-\frac{x-2}{x+3}

Multiply the first fraction by x+3 and the second by x+2;

\displaystyle B=(x+3)\frac{x-3}{(x+3)(x+2)}-(x+2)\frac{x-2}{(x+3)(x+2)}

Operating:

\displaystyle B=\frac{x^2-9}{(x+3)(x+2)}-\frac{x^2-4}{(x+3)(x+2)}

Subtracting both fractions with like denominators:

\displaystyle B=\frac{x^2-9-(x^2-4)}{(x+3)(x+2)}

Simplifying:

\displaystyle B=\frac{-5}{(x+3)(x+2)}

Thus the two expressions are:

\displaystyle A=\frac{x-2}{x+3}

And

\displaystyle B=\frac{-5}{(x+3)(x+2)}

8 0
3 years ago
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