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3 years ago

11

3x+2(4x+2)2(6x+1) what is the answer

2 answers:

Len [333]3 years ago

7 0

The answer is 65 I think

Semenov [28]3 years ago

3 0

The answer is x=2.....

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Three less than n equals 2 (solve for n)

ladessa [460]

If three less than 'n' equals 2, 'n'=5

3+2=5

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3 years ago

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rectangle ABCD is shown on the coordinate grid below. which of the following graphs represent the translation of rectangle ABCD

inn [45]

Answer:

the answer is c glad i could help

4 0

4 years ago

The rate of change is ____.<br><br>A.-4<br><br>B.-1<br><br>C.1<br><br>D.4

scZoUnD [109]

Answer: -4

Explanation: To find the rate of change, notice that the y values in the

table decrease by 4 each time and the x values increase by 1 each time.

Therefore, the rate of change, or the <em>change in y </em><em>/</em><em> change in x </em>is 4/-1 or -4.

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3 years ago

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A small company’s net income for the first six months of the year was $76,500 and for the last six months it was $100,000. What

kkurt [141]

Hello

Answer: The ratio of the first 6 months to the last 6 months is 0.765.
First, we need to determine the amount of money they had in each 6 month period.
In the first 6 months they had: 6 x 76500 = 459000
In the second 6 months they had: 6 x 100000 = 600000
Divide the 2 values to find the ratio: 459000 / 600000 = 0.765
Have a nice day

8 0

3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago