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3 years ago

Suppose x and y vary together such that x is 7 times as large as y. write a formula that defines x in terms of y.

Mathematics

1 answer:

Ainat [17]3 years ago

4 0

Since y is less than x by 7 times $\frac{x}{7}=y$ so in terms of x we write
x=7y

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16 POINTS!!! HELP ASAP

nadya68 [22]

Answer:

The answer is 358.01

Step-by-step explanation:

Given:

$3\times 10^{2}+5\times 10^{1}+8\times 10^{0}+1\times \frac{1}{10^{2}}$

Now, to solve first we crack the power notation then multiply and then do the addition:

$3\times 100+5\times 10+8\times 1+1\times \frac{1}{100}$

(The value of $10^{1}$ is 10 and the value of $10^{0}$ is 1.)

$=300+50+8+\frac{1}{100}$

$=300+50+8+0.01$

$=358.01$ .

Therefore, the answer is 358.01 .

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aniked [119]

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3 years ago

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WARRIOR [948]

Answer:

Step-by-step explanation:

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3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

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s2008m [1.1K]

Answer:

I think 2. is A i dont know the other two

Step-by-step explanation:

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3 years ago