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2 years ago

√2 x √27 use the properties of radicals to simplify

Mathematics

1 answer:

andre [41]2 years ago

7 0

Answer: look at the image

Step-by-step explanation:

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Cherries are on sale for $16 for 4 pounds. Jordyn pays $88.00 for cherries to make a cobbler. How many pounds of cherries doses

Tamiku [17]

She bought 5.5 pounds of cherries

explanation-
88 divided by 16 would be 5.5

7 0

2 years ago

Solve the proportion. i need help.<br> x+3/6=8/3<br><br> x=?

stellarik [79]

Answer:

x = 2 1/6 (13/6)

Step-by-step explanation:

to solve the proportion you must first find the LCF or least common factor

that is 6

so turn 8/3 into 16/6 by multiplying it by 2/2

x + 3/6 = 16/6

then you subtract 3/6 from both sides

x = 13/6

x is 2 1/6

6 0

2 years ago

Read 2 more answers

Which measure is greater?? One inch or one centimeter

Nimfa-mama [501]

One inch is the greater measure

8 0

3 years ago

A book sold 31,700 copies in its first month of release. Suppose this represents 8.8% of the number of copies sold to date. How

elena-14-01-66 [18.8K]

For this case we propose a rule of three:

31700 ------------> 8.8%

x ----------------------> 100%

Where "x" represents the number of copies sold to date.

$x = \frac {100 * 31700} {8.8}\\x = \frac {3170000} {8.8}\\x = 360,227.273$

Thus, to date 360,227 copies of the book have been sold.

ANswer:

To date, 360,227 copies of the book have been sold.

8 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago