All categories

3 years ago

In which quadrant of the coordinate graph does point F lie?

Mathematics

2 answers:

cupoosta [38]3 years ago

5 0

4 buddy now don’t forget that Brainly answer

irga5000 [103]3 years ago

5 0

Answer:

Quadrant IV

Step-by-step explanation:

use this image for future reference! hope it helps ;)

You might be interested in

50+x(greater than or equal to)300

nikdorinn [45]

Answer:

x ≥ 250

Step-by-step explanation:

Note that "greater than or equal to" looks like ≥

50 + x ≥ 300

Isolate the variable. What you do to one side, you do to the other. Subtract 50 from both sides.

x + 50 (-50) ≥ 300 (-50)

x ≥ 300 - 50

x ≥ 250

x ≥ 250 is your answer.

4 0

3 years ago

The donaldson's swimming pool measures 15 metters long 8 meters wide and 3 meters deep how many cubic meters if water will the p

Svetradugi [14.3K]

V=l*w*h

V= 15*8*3

V= 120*3

V= 360

The pool will hold 360 cubic meters.

7 0

4 years ago

What is the solution to the system of equations? <br>y=2/3x+3<br>x=-2

arsen [322]

Answer:

(x, y) and (2, 49)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

6.7a - 4.8a = ???<br> help if u can.

GrogVix [38]

Answer:

1.9a

Step-by-step explanation:

6.7a - 4.8a = 1.9a

7 0

3 years ago

Read 2 more answers

<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

4 years ago