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2 years ago

6

A table of values of a linear function is shown below. Find the output when the input is n. Type your answer in the space provid

ed.

1 answer:

Mumz [18]2 years ago

3 0

It’s correct answer believe me

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A dance academy charges $24 per class and a one-time registration fee of $15. a student paid a total of $687 to the academy. fin

Georgia [21]

28 is the correct answer

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4 years ago

5. Decide whether each equation is true for all, one, or no values of x.

S_A_V [24]

C doesn’t have a value , x equals 0 , x=0

6 0

3 years ago

kodGreya [7K]

Answer:

$24.35

Step-by-step explanation:

We will use the compound interest formula provided to solve this problem:

$A=P(1+\frac{r}{n} )^{nt}$

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

<em />

First, change 1% into a decimal:

1% -> $\frac{1}{100}$ -> 0.01

Since the interest is compounded monthly, we will use 12 for n. Lets plug in the values now:

$A=800(1+\frac{0.01}{12})^{12(3)}$

$A=824.35$

Lastly, subtract <em>A </em>from the principal to get the interest earned:

$824.35 - 800 = 24.35$

8 0

3 years ago

Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 15 samples are studied, and they can be cons

stiks02 [169]

Answer:

C) At most one sample is mutated

Step-by-step explanation:

If there are 15 samples, it means that 15 is the total (100%) of samples. Then, if we know that there is a chance that 3% are mutated, then we calculate the 3% of 15:

$Mutated= \frac{3. 15}{100} = 0.45$

This means that at most one sample is mutated, as this result is not zero (we discard answer A), and 0.45 is not more than half of the samples.

5 0

3 years ago

Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.

skad [1K]

DfLet $f(x)=\sec x$ and $g(x)=\sqrt x$ . Then

$y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)$

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

where $u=g(x)=\sqrt x$ , so that $y=f(g(x))=f(u)=\sec u$ . We have

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}$
$\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}$

5 0

3 years ago