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3 years ago

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If , f(x)= 3/(x+2) minus the square root of x-3, complete the following statement: f(19) = _____ (need help as soon as possible

not sure if the answer is just 19?)

2 answers:

Svet_ta [14]3 years ago

4 0

All you need to do is plug 19 in for x and solve.

f(19) = 3/(19+2) - sqrt(19-3) = 1/7 - sqrt(16)= 1/7 - 4

The answer is -3.8571 or as a fraction -27/7

Natasha_Volkova [10]3 years ago

4 0

Answer:

$f(x)= \frac{3}{x+2}- \sqrt{x-3}$

the given function has domain

x+2≠0

x≠ -2

x-3≥0

x≥3

That is , x ∈ [3,∞]

Now, calculating ,f (19)

$f(19)= \frac{3}{19+2}-\sqrt{19-3}\\\\ = \frac{3}{21} - \sqrt{16}\\\\ f(19)=\frac{1}{7}-4\\\\=\frac{1-28}{7}\\\\=\frac{-27}{7}$

$f(19)=-[3\frac{6}{7}]$

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What is the equation of the line that is parallel to the given

finlep [7]

Answer

O 4x+3y=-6

Step-by-step explanation:

Hello, I can help you witht this,

to solve this you need

step one

the line we are looking for is paraller to the line in the graph, it means both of then have the same slope.

find the slope of the line in the graph

remember

to find the slope of a line with two known points you can use

Let

$P1(x_{1},y_{1})}\\P1(x_{2},y_{2})\\slope\\s=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

pick two points from he graph

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P2(3,-1)

put the values into the equation

$slope(m)\\m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\m=\frac{-1-3}{3-0}\\m= \frac{-4}{3}$

so, the slope of the line we are looking for is m=-4/3 and passes through the point (-3,2)

step 2

use the equation

$y-y_{1}=m(x-x_{1})$

to ding the equation of the line

P1(-3,2)

$y-y_{1}=m(x-x_{1})\\y-2=\frac{-4}{3} (x+3})\\isolate\ y\\y-2=\frac{-4x}{3}-4\\y=\frac{-4x}{3}-4+2\\y+\frac{4x}{3}=-2\\multiply\ each\ side\ by\ 3\\3(y+\frac{4x}{3})=3*-2\\\\3y+4x=-6\\4x+3y=-6\\$

and that's all, that is the equation of the line that is parallel to the given

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Have a good day.

7 0

3 years ago

Solve this please show your work <3

Nookie1986 [14]

Given:

The expression is

$\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=$

To find:

The simplified form of the expression.

Solution:

We have,

$\sqrt[3]{48}=\sqrt[3]{8\cdot \_\_}=$

The expression $\sqrt[3]{48}$ can be written as

$\sqrt[3]{48}=\sqrt[3]{8\cdot 6}$

$\sqrt[3]{48}=\sqrt[3]{8}\cdot \sqrt[3]{6}$ $[\because \sqrt[3]{ab}=\sqrt[3]{a}\sqrt[3]{b}]$

$\sqrt[3]{48}=2\cdot \sqrt[3]{6}$

$\sqrt[3]{48}=2\sqrt[3]{6}$

Therefore, $\sqrt[3]{48}=\sqrt[3]{8\cdot 6}=2\sqrt[3]{6}$ .

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2 years ago

What two numbers multiply to 6 & add up to 8

Monica [59]

Answer: 2+4=6, and 2*4=8.

Step-by-step explanation:

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2 years ago

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A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 99% c

Lana71 [14]

Answer:

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Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What is the minimum sample size required to create the specified confidence interval

This is n when $M = 0.06, \sigma = 1.1$

$0.06 = 2.575*\frac{1.1}{\sqrt{n}}$

$0.06\sqrt{n} = 2.575*1.1$

$\sqrt{n} = \frac{2.575*1.1}{0.06}$

$(\sqrt{n})^{2} = (\frac{2.575*1.1}{0.06})^{2}$

$n = 2228.6$

Rounding up

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