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alukav5142 [94]
3 years ago
12

What can you conclude about the distance from zero for both and integer and its opposite

Mathematics
1 answer:
ruslelena [56]3 years ago
4 0
It is the absolute value
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What is an equation of the line with slope 2 that goes through point (3/2, 6)
Anna007 [38]

The answers are in slope-intercept form, so you can see that the slope (coefficient of x) is only correct for choices b and c.

When you substitute x=3/2 into those equations, you find that b gives y=6 (correct) and c gives y=0 (incorrect).

The appropriate choice is ...

... b.) y = 2x+3

4 0
3 years ago
Can you answer this math homework? Please!
Savatey [412]

Answer:

1. C -> Marty's with a slope of 1/3

2.C -> 2x + 5y = -15

3. C -> y = 1/4x + 2

4. A -> x - 4y = 8

5. D -> 5

4 0
3 years ago
How do I do this a second way? I totally forgot because I haven't done it in so long.
ira [324]
You take away the 4 from the 8 which is equal to 4 then you have to barrow from the 2 to make the 4 to 14 and when you    wait you are basically right your calculation is right
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4 years ago
Carmine buys 8 plates for $1 each. He also buys 4 bowls. Each bowl costs twice as much as each plate. The store is having a sale
iris [78.8K]

Answer:

$13

Step-by-step explanation:

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7 0
3 years ago
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P³ = 1/8 please help me if anybody can .
ivanzaharov [21]

Answer:

p= \dfrac{1}{2}

Step-by-step explanation:

Given equation:

p^3=\dfrac{1}{8}

Cube root both sides:

\implies \sqrt[3]{p^3}= \sqrt[3]{\dfrac{1}{8}}

\implies p= \sqrt[3]{\dfrac{1}{8}}

\textsf{Apply exponent rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}}:

\implies p= \left(\dfrac{1}{8}\right)^{\frac{1}{3}}

\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

\implies p= \dfrac{1^{\frac{1}{3}}}{8^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad 1^a=1:

\implies p= \dfrac{1}{8^{\frac{1}{3}}}

Rewrite 8 as 2³:

\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}

Simplify:

\implies p= \dfrac{1}{2^{\frac{3}{3}}}

\implies p= \dfrac{1}{2^{1}}

\implies p= \dfrac{1}{2}

3 0
1 year ago
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