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ra1l [238]
3 years ago
15

Explain what the scale factor for each category means

Mathematics
1 answer:
Rina8888 [55]3 years ago
4 0
The scale factor for math numbers are the numbers you multiply to get an anwser for its factor.
Ex: factor of 2 is 2,4,6,8 and so on with each number
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What is the x-intercept of the line in the equation 4x+2y=12
Fynjy0 [20]
I think that x=2 and so does y=2. Hope that answers your question.
3 0
3 years ago
Read 2 more answers
5x+4.27xy+20y+36y-19x+9.11
mote1985 [20]
So what I did was the terms that were alike like this: (5x -19x)+(20y+36y)+4.27xy+9.11. Then I simplified and got -14x+56y+4.27xy+9.11. The terms 4.27xy and 9.11 were not in ()'s because they were the only ones with an alike term. Hope that helps and also try to doing the steps and see what you get. If you get it wrong look to see where you messed up and try again! If I am wrong please notify me, I would appreciate it! 
6 0
3 years ago
Test 4,580 for divisibility by 2, 3, 5, 9, and 10.
Hatshy [7]

Answer:

Find the X and Y Intercepts 3x-5y=-20

Find the X and Y Intercepts 3x-5y=-20. 3x−5y=−20 3 x - 5 y = - 20. Find the x-intercepts. Tap for more steps

Step-by-step explanation:

5 0
2 years ago
Convert 50 kilogram to pounds. (The conversion factor from kilogram to pound is 2.2046.) A. 52.2 lb. B. 110.2 lb. C. 22.6 lb. D.
ss7ja [257]

Answer: (B) = 110.2

50*2.2046 = 110.23

5 0
3 years ago
Consider the region bounded by the curves y=|x^2+x-12|,x=-5,and x=5 and the x-axis
Tasya [4]
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

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}

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}

the last one you can do yourself, it is \frac{50}{3}
the sum is \frac{23}{6}+\frac{343}{6}+\frac{50}{3}=\frac{233}{3}


so the area under the curve is \frac{233}{3}
6 0
2 years ago
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