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3 years ago

The table and the graph below each show a different relationship between the same two variables, x and y:

Mathematics

1 answer:

gavmur [86]3 years ago

7 0

The value of y would be 20 times more valuable than 12 because the pattern for this table is y=20x. Let me know if I'm right :)

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Which expression is equivalent to 2(x+7)-18+4/5 ?

Vadim26 [7]

Answer:

4.7/5

Step-by-step explanation:

7 0

3 years ago

Why do you add a zero when multiplying two digits by two digits

malfutka [58]

When you add a zero to the problem, it shows you are now multiplying in the double-digits. Let's say you are multiplying 12x12. First you cover up the one then multiply them. Then cover up the two and you have 10. When you multiply them, it ends in a zero. Don't think of it as adding a zero but adding to products.

6 0

3 years ago

Alabama papas must average 80 on five test . Scores on the first four test were 81 73 84 and 76 what is the lowest score she can

Salsk061 [2.6K]

She has to score at least 86 on her fifth test.

Further explanation:

We will use the average formula to find the fifth value.

Given

$Average\ required=80\\x_1=81\\x_2=73\\x_3=84\\x_4=76\\x_5=?$

We have to find x5

The formula for average is:

$x=\frac{Sum\ of\ values}{Total\ number\ of\ values}\\Here,\\n=5\\So,\\x=\frac{x_1+x_2+x_3+x_4+x_5}{5}\\Putting\ values\\80=\frac{81+73+84+76+x_5}{5}$

$80*5=314+x_5\\400=314+x_5\\x_5=400-314\\x_5=86$

She has to score at least 86 on her fifth test.

Keywords: Average, Word Problems

Learn more about average at:

#LearnwithBrainly

7 0

4 years ago

$850 flat fee for 31 to 50 guests

AlladinOne [14]

Answer:

gdhcskjsopqdw0fi9sdahjqa

Step-by-step explanation:

you stupio person you told on me your deab

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%20%5Cfrac%7Bdx%7D%7Bxln%5E%7Bp%7Dx%20%7D" id="TexFormula1" title=" \int \frac{dx}{xl

Eva8 [605]

Substitute $u=\ln x$ , so that $\mathrm du=\dfrac{\mathrm dx}x$ . The integral is then equivalent to

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p\neq1\\\\\ln|u|+C&\text{for }p=1\end{cases}$

Then transforming back to $x$ gives

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p\neq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}$

4 0

3 years ago