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12345 [234]
3 years ago
9

John needs to make a scale drawing of his school building for art class. If the building is 256.25 meters long, and John scales

it down using a ratio of 25 meters to 1 inch, how long will the building be in the sketch?
12.62 inches
Mathematics
1 answer:
Neporo4naja [7]3 years ago
6 0
The answer is 10.25 inches
256.25/25= 10.25 inches. hope this helped

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aleksandrvk [35]

okokok

Step-by-step explanation:

5 0
3 years ago
maya tried to play a new video game.on her first 4 tries,she lost 14 points,lost 8 points,and finnaly gave up.write and find the
Pavel [41]
-22 points is what she lost

4 0
3 years ago
Read 2 more answers
Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8.
Oxana [17]

Answer:

6m + n + 2

Step-by-step explanation:

[7m + 3n - 11 - 2m] + [n + 5 - -m - 3n + 8]

[5m + 3n - 11] + [n + 5 + m - 3n + 8]

5m + 3n - 11 + (-2n + 13 + m)

5m + 3n - 11 - 2n + 13 + m

6m + n + 2

3 0
3 years ago
Substitute t=3 and t=5 to determine if the two expressions are equivalent.
motikmotik

Answer:

E

because both sides are equal when substituted by 3 & 5

4 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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