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Citrus2011 [14]
3 years ago
10

Help me please free cookies if you help ✋

Mathematics
2 answers:
fomenos3 years ago
7 0

Answer:

answer is 0.19in^3

Step-by-step explanation:

i would like the cookie now...

tensa zangetsu [6.8K]3 years ago
6 0
Choclate Chip Cookies?
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A trapezoid has 63.4 kilometers base and 23.2 kilometers base, as well as 75.3 kilometers height. What is the area?
Finger [1]
Make sure to multiply everything together, then check it with a calculater.
Hope this works:)
7 0
3 years ago
Read 2 more answers
1-4
Oksi-84 [34.3K]

Answer:

282 Hours after 6 weeks

Step-by-step explanation:

Multiply the 6 weeks and the 48 hour a week adverage and theres your answer!

3 0
2 years ago
Can someone help me please.
zhannawk [14.2K]

Answer:

  1. Lena's Method: 280
  2. Jackson's Method: 280

Step-by-step explanation:

Lena's Method:

  1. 8 + 12 = 20
  2. Plug 20 in: 14(20)
  3. 14(20) = 14 × 20 = 280

Jackson's Method:

  1. 14 × 8 = 112
  2. 14 × 12 = 168
  3. Re-write the expression: 112 + 168
  4. 112 + 168 = 280

I hope this helps!

8 0
3 years ago
Which of these is an example of the Commutative property of addition? And Explain why? 3+5=4+4 or 3+5=5+3
mestny [16]

Answer:

3+5=5+3

Step-by-step explanation:

Because the commutative property means 'flip-flop'

Hope this helps! = )

5 0
2 years ago
Read 2 more answers
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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