All categories

2 years ago

The difference between the square roots of a number is 30. What is the number?

Mathematics

1 answer:

Vilka [71]2 years ago

5 0

The difference between the square roots of a number 30 is 15 I am pretty sure it is 15

You might be interested in

Bridgette measured an Italian restaurant and made a scale drawing. The scale of the drawing was 8 millimeters: 5 meters. The res

Goshia [24]

Answer:

The kitchen drawing is 48 millimeters long.

7 0

2 years ago

Find the slope of the line containing points (1,3) (-3,-8)

Dmitry_Shevchenko [17]

Y=mx+c
1/-3=m3/8+c
m= 11/4

4 0

3 years ago

16/__ = -8<br> ___/12 = -6

Karolina [17]

Answer:

1.) - 2

2.) - 72

Step-by-step explanation:

hope this helps

6 0

3 years ago

Read 2 more answers

Bo paid bills with 1/2 of his paycheck and put 1/5 of the remainder in savings. The rest of his paycheck was divided equally amo

Alecsey [184]

If you covert it to 18/30 then it is much simpler yet if you still need help then read on

4/30

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5C%20%28x%5E3-6x%5E2%2B9x%2B3%29%283x%5E2-12x%2B9%29%20dx" id="TexFormula1" title="\int

AlexFokin [52]

If the integral is simply

$\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx$

then notice that

$\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx$

which means you can compute the integral easily with a substitution

$u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx$

Under this transformation, the integral is

$\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}$

On the other hand, in case you're missing a symbol and the integral is actually

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx$

then first carry out the division:

$\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}$

Now, $3x^2-12x+9=3(x-3)(x-1)$ , so to integrate the remainder term you can decompose it into partial fractions:

$-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}$

$9-2x=a(x-1)+b(x-3)$

$x=1\implies7=-2b\implies b=-\dfrac72$

$x=3\implies3=2a\implies a=\dfrac32$

$\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}$

Then the integral would be

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}$

which can be rewritten in several ways, such as

$\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C$

6 0

3 years ago