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Svetradugi [14.3K]
2 years ago
14

Which expression can be used to solve the problem below? On a recent math test,

Mathematics
2 answers:
lora16 [44]2 years ago
3 0

Answer:

4

Step-by-step explanation:

Because she gets 3 per multiple choice question, so 3 times 18

she gets 5 per response question, so 5 times 6

lys-0071 [83]2 years ago
3 0
The expression you would use would be 4.
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Diameter of a circle is two units. What is the radius of the circle?
bonufazy [111]
Answer is 3.5 feet enjoy
5 0
2 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
Help no time plzzzz plz plz​
Sav [38]

Answer:

FALSE

Step-by-step explanation:

<E in ∆AED ≅ <E in ∆CEB.

Both are 90°.

Side ED ≅ Side EB

Side AD ≅ Side CB.

Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.

8 0
2 years ago
Answer this fast please
Likurg_2 [28]

Answer:

negative

Step-by-step explanation:

b is a negative number because it is less than  -1, so b/5 is a negative number because negative/positive = negative

6 0
2 years ago
Read 2 more answers
The first five terms of a linear sequence are given below. 7 , 12 , 17, 22 , 27 , ... What is the next term of the sequence?
Norma-Jean [14]

Answer:

<h2>32</h2>

Step-by-step explanation:

a_1=7\\a_2=7+5=12\\a_3=12+5=17\\a_4=17+5=22\\a_5=22+5=27\\\\a_6=27+5=32\\a_7=32+5=37\\a_8=37+5=42\\a_9=42+5=47\\\vdots

8 0
2 years ago
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