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Nina [5.8K]
3 years ago
6

Please help me solve 3x^3 - 4x^2 + 9x - 12. I am trying to factor this completely.

Mathematics
1 answer:
nikklg [1K]3 years ago
5 0
It’s already factored
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How do you verify this trigonometric identity? cos^2 θcot^2 θ = cot^2 θ - cos^2 θ
sattari [20]
<h3>Explanation:</h3>

Replace cos^2(θ) with 1-sin^2(θ), and cot(θ) with cos(θ)/sin(θ).

  cos^2(θ)cot^2(θ) = cot^2(θ) - cos^2(θ)

  (1 -sin^2(θ))cot^2(θ) =  . . . . . replace cos^2 with 1-sin^2

  cot^2(θ) -sin^2(θ)·cos^2(θ)/sin^2(θ) = . . . . . replace cot with cos/sin

  cot^2(θ) -cos^2(θ) = cot^2(θ) -cos^2(θ) . . . as desired

8 0
3 years ago
HELP PLZ When x = -2, which of the following points is on the graphed line y = -4x + 3?
trapecia [35]

y = f(x) =  - 4x + 3 \\ \Rightarrow f( - 2) =  - 4( - 2) + 3 = 8 + 3 = 11 \\ \Rightarrow ( - 2,11)

6 0
3 years ago
Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0
3 years ago
Question
ANTONII [103]

The approximate area of the park on the grid is: E. about 40 km² to 50 km².

<h3>How to Find the Approximate Area on a Coordinate Grid?</h3>

The number of square on a coordinate grid that is covered determines the area covered. We can make an estimate by counting how many of this square on the coordinate grid that is covered, then find out the area depending on how much square area each grid represents.

In the coordinate plane given, which shows a park, we are told that each of the square on the grid equals 1 k = square kilometer.

The number of each of these squares we can find that is covered by the park on the grid is: 48 squares.

Therefore, the area of 48 squares on the grid = 48 × 1 = 48 km². Since not all squares are fully covered by the park, we can state that the approximate area of the park on the grid is: E. about 40 km² to 50 km².

Learn more about the approximate area on a grid on:

brainly.com/question/368985

#SPJ1

7 0
1 year ago
There are "Brainly Bots" giving links. Do not click them. You can tell if it says something about an answer with a link to a lin
Lerok [7]

Thanks for letting me know! I have been encountering those bots lately when I am looking for an answer to my question. We need more people like you in the community so thanks for letting others know! We appreciate it!!!

3 0
3 years ago
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