Which expression is equivalent to 4(x + 1) - 7x?
2 answers:
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Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-5%7D_%7B-4%7D%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-5%7D_%7B-4%7D%3D%28%5Cfrac%7B-125%7D%7B3%7D%2B%5Cfrac%7B25%7D%7B2%7D%2B60%29-%28%5Cfrac%7B64%7D%7B3%7D%2B8%2B48%29%3D%5Cfrac%7B23%7D%7B6%7D)
next one
![\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%3D-1%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-4%7D_%7B3%7D%3D-1%28%28-64%2F3%29%2B8%2B48%29-%289%2B%289%2F2%29-36%29%29%3D%5Cfrac%7B343%7D%7B6%7D)
the last one you can do yourself, it is
the sum is
so the area under the curve is
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
next one
the last one you can do yourself, it is
the sum is
so the area under the curve is
Answer:
The area and the perimeter of the picture are:
- <u>Area = 160 cm^2</u>
- <u>Perimeter = 67.31 cm</u>
Step-by-step explanation:
To find the area of that figure, you can find the area how if it was a rectangle and next subtract the area of the triangle in the upper part. The area of a rectangle could be found by the next formula:
- Area of a rectangle = base * height
As you can see in the picture, the base is 16 cm and the height is 12 cm, then we replace in the formula:
- Area of a rectangle = 16 cm * 12 cm
- Area of a rectangle = 192 cm^2
Now, we calculate the area of the triangle to subtract to the area we found and obtain the real area, the formula to obtain the area of a triangle is:
- Area of a triangle = (base * height) / 2
The height of the triangle is 8 cm, and the base is 8 cm too, because you subtract to the base of the rectangle (16 cm) the measurements in the upper part (16 - 4 - 4 = 8), Now, we replace in the formula:
- Area of a triangle = (8 cm * 8 cm) / 2
- Area of a triangle = (64 cm^2) / 2
- Area of a triangle = 32 cm^2
We subtract to the found area:
- Area of the picture = 192 cm^2 - 32 cm^2
- <u>Area of the picture = 160 cm^2</u>
To find the perimeter, you must add all the sides of the picture, but, as you can see, there is a side that doesn't have the measurent, this is the hypotenuse of the triangle used before, but how we know the other sides, we can use Pythagorean theorem:
Where:
- a = Opposite leg (8 cm)
- b = Adjacent leg (8 cm)
So, we replace in the theorem:
(and we clear c)
-
- c = 11.3137085 cm
- c ≅ 11.31 cm
At last, we add all the sides of the picture begining by the base and going by the left side:
- Perimeter of the picture = 16 cm + 12 cm + 4 cm + 11.31 cm + 8 cm + 4 cm + 12 cm
- <u>Perimeter of the picture = 67.31 cm approximately</u>.