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3 years ago

Given f(x)=-4x+7 find f(2) how do i slove this

Mathematics

1 answer:

yan [13]3 years ago

8 0

Answer:

-1

Step-by-step explanation:

given f(x)=-4x+7 find f(2)

f(2) means question : if x=2, f(2)=?

f(2)= -4x2+7=-1

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Lines m and p are perpendicular if the slope of line is 1, 25th what is the slope of line p

Rainbow [258]

-1 is the opposite reciprical (perpendicular)

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3 years ago

Fifty trees grow in a forest. Workers cut down 23 of the trees. Then they plant 38 new ones. After all the cutting and planting,

ololo11 [35]

Answer:

Option A. 65 trees

Step-by-step explanation:

From the question given:

50 trees was in the forest. Then 23 of them were cut down.

The remaining trees = 50 — 23 = 27 trees.

Now 38 new trees were plant.

Total trees after cutting and planting = 27 + 38 = 65 trees

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3 years ago

Read 2 more answers

Find the future value of 575 at 5.5% compounded quarterly for 5 years. Round to the nearest cent

Readme [11.4K]

Answer:

Future value = $755.61 ( to the nearest cent)

Step-by-step explanation:

The formula for calculating the future value of an invested amount compounded periodically for a number of years is given as:

$FV = PV (1+\frac{r}{n} )^{n*t}$

where:

FV = future value = ???

PV = present value = $575

r = interest rate in decimal = 5.5% = 0.055

n = number of compounding periods per year = quarterly = 4

t = time of investment = 5 years

∴ $FV = 575 (1+\frac{0.055}{4} )^{4*5}$

$FV = 575 (1+0.01375)^{20}\\FV = 575 (1.01375 )^{20}\\FV = 575 * 1.3141\\FV = 755.607$

∴ Future value = $755.61 ( to the nearest cent)

4 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.)

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3 years ago

jeff volunteers his time by working at an animal shelter. each year he works for total of 240 hours. so far this year, he has wo

aleksandrvk [35]

Answer:

h=240-97

Step-by-step explanation:

5 0

2 years ago