Three more than the product of six and a number, increased by nine in the simplest form

Mathematics

1 answer:

Nookie1986 [14]3 years ago

4 0

Answer: 12+6x

Step-by-step explanation:

three more means addition +3

the product of 6 and a number means multiplying 6 by a variable 6x

increased by 9 means addition +9

put it together and we have 3+6x+9

now we combine like terms

3+9+6x

12+6x

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Simplify (19x2 + 16x + 45) - (9x2 + 13x + 23).

Yuri [45]

19x2-9x2= 10x2
16x-13x= 3x
45-23= 22

(10x2+3x+22)

8 0

3 years ago

How do you do this question?

Andrei [34K]

Answer:

Solution : Option B, or 9π

Step-by-step explanation:

We are given that y = x, x = 3, and y = 0.

Now assume we have a circle that models the given information. The radius will be x, so to determine the area of that circle we have πx². And knowing that x = 3 and y = 0, we have the following integral:

$\int _0^3$

So our set up for solving this problem, would be such:

$\int _0^3x^2\pi \:$

By solving this integral we receive our solution:

$\int _0^3x^2\pi dx,\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> \pi \cdot \int _0^3x^2dx\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\=> \pi \left[\frac{x^{2+1}}{2+1}\right]^3_0\\=> \pi \left[\frac{x^3}{3}\right]^3_0\\\mathrm{Compute\:the\:boundaries}: \left[\frac{x^3}{3}\right]^3_0=9\\\mathrm{Substitute:9\pi }$