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3 years ago

X=5 y = 3, x² + 3(x +y)= HELP FAST

Mathematics

1 answer:

umka2103 [35]3 years ago

5 0

9514 1404 393

Answer:

Step-by-step explanation:

Put the variable values in place of the corresponding variables, and do the arithmetic.

x² +3(x +y) =

5² +3(5 +3) = 25 +3(8) = 25 +24 = 49

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Identify the initial value and rate of change for the graph shown. (4 points)

fiasKO [112]

Answer:

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Step-by-step explanation:

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2 years ago

How do I write a compound inequality?

ss7ja [257]

Answer:

it can either be written as x > -1 and x < 2 or as -1 < x < 2.

Step-by-step explanation:

The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. A number is a solution to the compound inequality if the number is a solution to both inequalities. It can either be written as x > -1 and x < 2 or as -1 < x < 2.

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3 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$

6 0

3 years ago

F(x)= x^2– 3x + 9<br> g(x) = 3x^3+ 2x^2– 4x – 9<br> Find (f - g)(x).

denis-greek [22]

Answer:

$\large \boxed{\sf \ \ -3x^3-x^2+x+18 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

$(f-g)(x)=f(x)-g(x)=x^2-3x+9-(3x^3+2x^2-4x-9)\\\\=x^2-3x+9-3x^3-2x^2+4x+9\\\\=\boxed{-3x^3-x^2+x+18}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

3 years ago

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data

Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

The interval will be

$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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3 years ago