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garri49 [273]
3 years ago
11

Can you answer this please

Mathematics
1 answer:
sertanlavr [38]3 years ago
4 0

Answer:

Hi, there your answer is y=3x+4

Step-by-step explanation:

I did rise over run

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What is the full question ?
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Find the domain and range of the relation.
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the answer is in the picture above

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2 years ago
Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))
Marrrta [24]

Answer:

\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Step-by-step explanation:

\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine \frac{1}{c} + \frac{1}{h}

\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}

Combine the bottom, too.

=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}

Apply the fraction rule

=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}

Cancel

=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

Therefore, \frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}

5 0
3 years ago
A survey was conducted to determine the amount of time, on average, during a given week SCAD students spend outside of class on
fiasKO [112]

Answer:

Standard Deviation = 5.928

Step-by-step explanation:

a) Data:

Days  Hours spent  (Mean - Hour)²

1              5                61.356

2             7                34.024

3            11                  3.360

4           14                   1.362

5           18               26.698

6          22               84.034

6 days 77 hours,    210.834

mean

77/6 = 12.833    and 210.83/6 =  35.139

Therefore, the square root of 35.139 = 5.928

b) The standard deviation of 5.928 shows how the hours students spend outside of class on class work varies from the mean of the total hours they spend outside of class on class work.

5 0
3 years ago
Can someone please help me find the area of this parallelogram.
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Answer:

72

Step-by-step explanation:

18 + 18 + 6 + 6 + 12 + 12

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