Help me please thanks :)

What are the coordinates of the origin? A) (0, 1) B) (x, y) C) (0, 0) D) (1, 0)

ratelena [41]

Answer:

i believe it's c

Step-by-step explanation:

hope this helps :)

4 0

3 years ago

Factor 9x^2+66xy+121y^2

Korvikt [17]

Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

[ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.

7 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$