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miv72 [106K]
3 years ago
5

Richard took a math quiz last week. There were 100 problems on the quiz and Richard answered 30% of them correctly. How many pro

blems did Richard get correct?
Mathematics
2 answers:
densk [106]3 years ago
8 0

Answer:

Step-by-step explanation:

100×3=300÷100

he got 3 correct.

kolbaska11 [484]3 years ago
6 0

Answer: he got 30 of them correctly

Step-by-step explanation: because 30% of 100 is 30.

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Sandra needs 6 pieces of rope, each measuring 8 inches in length. She is going to cut the 6 pieces from a rope that is 84 inches
soldier1979 [14.2K]
36 inches of rope will be left because 6*8 is 48 and you subtract 48 from 84 to get 36.
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Will give BRAINLIEST to the explanation that gives the most sense!!
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The numbers I have input, it’s seems like the answer is never.
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a class conduct a survey of 1000 students 6% of the students have forgotten their locker combinations how many students have for
solong [7]
This is a simple percentage question. 6/100 * 1000 = 60 people have forgotten
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3 years ago
4|m−n| if m=−7 and n=2 i really need to know what this is asap please
vekshin1

Answer:

36

Step-by-step explanation:

Plug in -7 as m and 2 as n into the expression:

4 | m - n |

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6 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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