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irinina [24]
3 years ago
9

Hey if anyone could help me with this it would be a big help!

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0

c is the answer i hope this help

Morgarella [4.7K]3 years ago
3 0

Answer:

sorry if I'm wrong but i think it's the first one 3.0 x 10^6

Step-by-step explanation:

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Using the equation y=kx and the table below, find the constant of proportionality (k).
elena-s [515]

Answer:

k = 9/4

Step-by-step explanation:

To find k, use the equation

y = kx

Pick a point

72 = k *32

Divide each side by 32

72/32 = k

9/4 = k

4 0
3 years ago
Read 2 more answers
Two numbers are greater than 10 but les than 40. Their GCF is 17. What are the numbers?
shusha [124]
The numbers are 17 and 34.
8 0
3 years ago
The vertex of this parabola is at (-3, 6). Which of the following could be its equation?
pishuonlain [190]
Since the vertex is at (-3, 6), the equation is given by
y = a(x + 3) + 6

Therefore, option A is the correct answer.
6 0
3 years ago
Solve using long division <br> Please
madreJ [45]

1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

Steps:

\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}

\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

Steps:

\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
3 years ago
(04.04)
harkovskaia [24]
I Think The answer is 35 units I hope it helps Message Me if I’m wrong and I’ll change My answer and fix it for you
6 0
4 years ago
Read 2 more answers
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