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Nady [450]
2 years ago
6

Someone help me out please

Mathematics
1 answer:
Reika [66]2 years ago
7 0

Answer:

Coordinates are

(1,-5)

(-6,1)

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I need help so bad this makes no sence
oksian1 [2.3K]
Math really can seem hopeless at times, can it? I can't recall the number of times that I've given up on math, but enough about me! There is a certain theorem in Geometry that states that the angle formed by two secant segments, 9x-5, is equal to half the difference between the intercepted arcs (64 and 158). I know this might sound extraterrestrial, but bear with me! Here is what it would look like in numerical form.

angle formed by secant segments = (1/2) (big intercepted arc - small intercepted arc)
(9x - 5) = (1/2)(158 - 64)
9x - 5 = (1/2)(94)
9x - 5 = 47
9x = 52
x = 52/9 or 5 7/9
6 0
3 years ago
At the rate 36 bottle caps in 10 seconds how many bottle caps can be made in 5min
Misha Larkins [42]
1,080 bottle caps. There's 60 seconds per minute so 36*6, Then that times 5 for the 5 minutes.
6 0
3 years ago
How would u solve these questions?
Tasya [4]
Can't see most of the pic
3 0
3 years ago
Product -6 and sum 1 what is the factors
den301095 [7]

Answer:

The answer is -2 & 3

Step-by-step explanation:

8 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
3 years ago
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