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Mekhanik [1.2K]
3 years ago
9

Yohanna is conditioning all summer to be fit for the next soccer season at his high school. He incorporated walking planks into

his daily workout regime. Each day, he will complete 4 more walking planks than the day before. If he starts with 5 walking planks on the first day, how many walking planks will he do on the twelfth day?

Mathematics
1 answer:
BARSIC [14]3 years ago
4 0

I'm pretty sure the answer is 53.

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which of the following is a polynomial expression a. 2^x + x + 3 b. x^2 + 2x + 7 or c. 1/x + 1/x2 + 1/x3 ( btw the ^2 means it’s
NeTakaya

(b)

A polynomial has the form

a_{0}x^{n} ± a_{1}x^{n-1} + ....

the only expression fitting this description is : x² + 2x + 7


5 0
3 years ago
14. Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,1) E (7,1) and F(1,5). Show your work.
Bess [88]
Hello : 
 the  <span>coordinates </span> circumcenter for ∆DEF : 
x = (1+7+1)/3
y = (1+1+5//3
x=3
y= 7/3
6 0
3 years ago
Suppose you have $500,000 for retirement in 20 years.Your account earns 6% interest. How much would you need to deposit in the a
laiz [17]

Answer:

30,000

Step-by-step explanation:

What I did was 6 divided by 100 and I got 0.06 times the the 500,000 and got 30,000 so your answer should be 30,000. Hope this helps and let me know if I'm wrong. have a great day!

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
2 years ago
Please help!
nasty-shy [4]
Right change flip it's simple
4 0
2 years ago
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