Ten is mistakenly written by Tim in question.
Gabriel is making a mixture of compost and soil. In which he wants to mix 2 part compost to 10 parts potting soil. And he wants to end up with ten kilogram
It means compost is 2 pars out of 12 and potting soil is 10 parts out of 12
so, (2 / 12) x 10 kilogram of compost is used
= 1.66667 kilogram of compost Gabriel should use.
Answer:14 m
Step-by-step explanation:
Dada
larga ![l=20\ m](https://tex.z-dn.net/?f=l%3D20%5C%20m)
superficie ![A=280\ m^2](https://tex.z-dn.net/?f=A%3D280%5C%20m%5E2)
Suponga que el ancho es w
La superficie está dada por el producto de la longitud y la anchura.
![\therefore A=lw\\\Rightarrow 280=20\times w\\\Rightarrow w=14\ m](https://tex.z-dn.net/?f=%5Ctherefore%20A%3Dlw%5C%5C%5CRightarrow%20280%3D20%5Ctimes%20w%5C%5C%5CRightarrow%20w%3D14%5C%20m)
Por lo tanto, el ancho es de 14 m.
Answer:
x=−3/4
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
435=−4x+285
Step 2: Flip the equation.
−4x+285=435
Step 3: Subtract 28/5 from both sides.
−4x+285−285=435−285
−4x=3
Step 4: Divide both sides by -4.
−4x−4=3−4
x=−34
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.
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx](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%2B%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%20%2B%20%5Cint%5Climits%5E3_5%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20)
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
![\frac{50}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B50%7D%7B3%7D)
the sum is
![\frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B23%7D%7B6%7D%2B%5Cfrac%7B343%7D%7B6%7D%2B%5Cfrac%7B50%7D%7B3%7D%3D%5Cfrac%7B233%7D%7B3%7D)
so the area under the curve is