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3 years ago

BRAINLIEST!!!

Mathematics

2 answers:

Papessa [141]3 years ago

4 0

For the equation, the rate of change would be four.
Now looking at the graph, look at each point of the line that hits perfectly on a grid. Count up four, over two. The rate of change for the graph would be 4/2, or 2. Hope this helps!

ankoles [38]3 years ago

4 0

Hi there!

Graph:
The rate of change is 2.

If we take the points (3,2) and (1, -2) and use slope triangles to find the slope, 4 ÷ 2 = 2, so the slope is 2.

Function:
The rate of change is 4.
The rage of change in a function in slope intercept form is always the number before x.

Hope this helps!

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If x/8 =5/4 then x=10 true false

galina1969 [7]

Answer:

true

Step-by-step explanation:

given

$\frac{x}{8}$ = $\frac{5}{4}$ ( cross- multiply )

4x = 40 ( divide both sides by 4)

x = 10

3 0

3 years ago

Juanita Domingo's parents want to establish a college trust for her. They want to make 16 quarterly withdrawals of $2000, with t

kotegsom [21]

Answer:

The amount to be deposited now to provide for this trust is $119,392.16.

Step-by-step explanation:

This problem is based on ordinary annuity.

An ordinary annuity is a sequence of fixed payments made, every consecutive period, over a fixed interval.

The formula to compute ordinary annuity is:

$OA=P[\frac{q^{n}-1}{q^{n}(q-1)}]$

Here <em>qⁿ </em>is:

$q^{n}=1+\frac{r}{Number\ of\ periods}=1+\frac{0.067}{4}=1.01675$

Compute the ordinary annuity as follows:

$OA=P[\frac{q^{n}-1}{q^{n}(q-1)}]=2000\times\frac{(1.01675)^{16}-1}{(1.01675)^{16}[1.01675-1]}=2000\times\frac{0.30445}{0.0051}=119392.16$

Thus, the amount to be deposited now to provide for this trust is $119,392.16.

4 0

3 years ago

Find The derivative of:(cosx/1+sinx)^3

forsale [732]

Answer:

$-\frac{3 \cdot cos^2x}{(1+sinx)^3}$

Step-by-step explanation:

$y = (\frac{cosx}{1+sinx})^3\\\\\frac{dy}{dx} = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{dy}{dx}(\frac{cosx}{1+sinx})$ $[ y = x^n\ \ \ => \ \ \ \frac{dy}{dx} = b \cdot x^{n-1} \ ]$

$= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{(1+sin x(-sinx) - cosx(cosx)}{(1+sinx)^2}\\\\$ $[\ \frac{u}{v} = \frac{v \dcot u'- u \cdot v'}{v^2}\ ]$

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