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Over [174]
3 years ago
6

What is the sum of 3(x + 3) and -2(x - 5)? A. X-2 B. x-1 C. X + 8 D. X + 19

Mathematics
1 answer:
Umnica [9.8K]3 years ago
7 0
The answer is actually d lol the idiot above me was right
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<span>4.23 times 9 equals 38.07 :)</span>
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Write 84 as a product of prime numbers
hoa [83]
Factors of 84: 1, 2<span>, </span>3<span>, 4, 6, </span>7<span>, 12, </span>14<span>, </span>21<span>, </span>28<span>, </span>42<span>, 84. Prime factorization: 84 = </span>2<span> x </span>2<span> x </span>3<span>x </span>7<span> which can also be written (</span>2^2<span>) x </span>3<span> x </span>7<span>.</span>
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4 years ago
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Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear
AlladinOne [14]

Answer:

x=\frac{83}{50}

Step-by-step explanation:

we know that

If the three points are collinear

then

m_A_B=m_A_C

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

step 1

Find the slope AB

we have

A(1,\frac{2}{3}),B(x,-\frac{4}{5})

substitute in the formula

m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}

m_A_B=\frac{\frac{-12-10}{15}}{x-1}

m_A_B=-\frac{22}{15(x-1)}

step 2

Find the slope AC

we have

A(1,\frac{2}{3}),C(-\frac{1}{2},4)

substitute in the formula

m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}

m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}

m_A_C=-\frac{20}{9}

step 3

Equate the slopes

m_A_B=m_A_C

-\frac{22}{15(x-1)}=-\frac{20}{9}

solve for x

15(x-1)20=22(9)

300x-300=198

300x=198+300

300x=498

x=\frac{498}{300}

simplify

x=\frac{83}{50}

8 0
4 years ago
If your cell phone bill was $67.85 and there is a 7.5% late fee how much will your bill be?
dedylja [7]

67.85 \times .075 = 5.08875 \\ 67.85 + 5.08875 = 72.94
bill is $72.94
6 0
4 years ago
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For what value of c will this function have one zero?<br> f(x) = X^2 +c
Phoenix [80]
X² + c is actually a quadratic function.

And x² + c = 0,  it usually has two zeros which are solutions.

But for when c = 0,

x² + c = 0

x² + 0 = 0

x² = 0

Taking the square root of both sides.

x = 0.  Here it only has one zero.

So the function x² + c,  only has one root for  c = 0.
6 0
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