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ruslelena [56]
2 years ago
7

Solve (6b^2+5)+(8b-20)

Mathematics
1 answer:
Lisa [10]2 years ago
7 0

The sum of the given expression is 6b^2+8b-15

<h3>Sum of expression</h3>

Given the following expression

(6b^2+5)+(8b-20)

We are to take the sum. On expansion;

6b^2+5 + 8b-20

Collect the like terms

6b^2+8b +5 - 20

6b^2+8b-15

Hence the sum of the given expression is 6b^2+8b-15

Learn more on sum of expression here: brainly.com/question/723406

#SPJ1

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Q(x)= 1/2x - 3; q(x) =-4. Find the value of x
Lelu [443]

Answer:

q(x) =-5

Step-by-step explanation:

I am solving this using Function Notation.

We are given:

q(x)= \frac{1}{2}x - 3

While we are to find:

q(x) =-4

In Function Notation, replace all the "x" you see with the value you are given.

So, in this case,

q(x)= \frac{1}{2}x - 3\\\\=\frac{1}{2}x - 3

Would become

q(x)= \frac{1}{2}x - 3\\\\=\frac{1}{2}(-4) - 3

Now we can solve.

\frac{1}{2}(-4) - 3

Use PEMDAS; Parenthesis first;

\frac{1}{2}(-\frac{4}{1})

Cross divide: 1 cancels 1, -4 ÷ 2 = -2

Hence, we are left with

(-2) - 3

-5

Therefore, the value of x is -5

8 0
2 years ago
What equation represents a circle whose center is located at (-6,2) and whose radius is 10 units?
Anna [14]
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So after substitution:
(x + 6)² + (y - 2)² = 100

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3 0
3 years ago
Write 7.63 × 10^–6 in an ordinary number.
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Answer:

0.00000763

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Answer:

15°

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Step-by-step explanation:

Since P is on the median of ΔABC, it is equidistant from points B and C as well as from C and Q. Thus, points B, C, and Q all lie on a circle centered at P. (See the attached diagram.)

The base angles (B and C) of triangle ABC are (180° -30°)/2 = 75°. This means arc QC of the circle centered at P has measure 150°. The diameter of circle P that includes point Q is defined to intersect circle P at R.

Central angle RPC is the difference between arcs QR and QC, so is 180° -150° = 30°. Inscribed angle RQC has half that measure, so is 15°.  Angle PQC has the same measure as angle RQC, so is 15°.

Angle PQC is 15°.

4 0
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