1/3 (3x+9) = 5-3 (x-2)

A group of friends goes out to lunch. The total bill, excluding tax, is $22. If tax on the meal is 6% and if a 20% tip is left o

TiliK225 [7]

Answer:

27.72

Step-by-step explanation:

Sorry about before i made a slight mistake

5 0

2 years ago

Find the volume of the figure if length = 1/2 m, width = 1/2 m and height = 6 meters. I WILL GIVE BRAINLIEST PLLZZZ SOMEONE HELP

goldfiish [28.3K]

Step-by-step explanation:

Volume of figure=length×width×height

=(1/2)×(1/2)×6

=0.5×3

=1.5m³

PLEASE GIVE BRAINLIEST

5 0

3 years ago

What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?

RideAnS [48]

Step-by-step explanation:

The function is as follows :

$f(x) = x^2 + 5x + 5$

We need to find the zeroes of the function in the simplest radical form.The zero of the above function is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here,

b = 5

a = 1

c = 5

So,

$x=\dfrac{-5\pm \sqrt{5^2-4\times 1\times 5} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{25-20} }{2(1)}\\\\x=\dfrac{-5\pm \sqrt{5} }{2}$

Hence, the correct option is (c).

6 0

3 years ago

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

as

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

so

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago

Estimate 4.7 % of 160 16 8 80 47

IRINA_888 [86]

Answer:

hhsjjsjsjsjsjsjsjsjsjsjjz

Step-by-step explanation:

jsjsjsjsjsjsjsjsjsjsjsjdjdjdjdj

5 0

3 years ago