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NemiM [27]
3 years ago
13

I need help plz answer

Mathematics
2 answers:
DiKsa [7]3 years ago
8 0

Answer:

12:4

3:1

18:6

Step-by-step explanation:

algol133 years ago
6 0

Answer:

when x is 12, y is 4

when x is 3, y is 1

when x is 18, y is 6

Step-by-step explanation:

Plug the value of x into the equation to solve.

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David played a video game 5 times. The list shows his scores.
viva [34]
You would add up all of the numbers and then divide them by 5 since you have a total of 5 different sets of numbers (To get the mean). Then you have to find the median by putting all of the numbers in numerical order and get the number in the middle. Finally you subtract the median from the mean and you end up getting “B” as your answer.
6 0
3 years ago
K = mv / 2 solve for m . HELP PLEASE HURRY!
Aleksandr-060686 [28]

9514 1404 393

Answer:

  m = 2k/v

Step-by-step explanation:

Identify the coefficient of m (v/2) and multiply by its inverse (2/v).

  (2/v)k = (2/v)m(v/2) . . . . multiply both sides of the equation by 2/v

  2k/v = m . . . simplify

  m = 2k/v

3 0
2 years ago
Given a monthly income of $1,500, how do you determine the yearly. salary this person will earn
Nina [5.8K]

Answer:

C: $1500 x 12

Step-by-step explanation:

It's simple. You've gotta multiply to get this answer.

7 0
3 years ago
how would a enter a payment that was accepted with a 4 month, 8% note in a payment of a $4800.00 account
Anon25 [30]

Answer:

Enter a payment of 5192.52.

Step-by-step explanation:

Consider the provided information.

The payment is $4800 with a 4 month, 8% note.

The amount can be calculated as:

Amount=p(1+\frac{r}{m})^{mt}

Where <em>p</em> is money invested, <em>r</em> is annual interest rate, <em>t</em> is number of years and <em>m</em> is number of period.

Now substitute p = 4800 r = 0.08 and m = 4 in the above formula.

Amount=4800(1+\frac{0.08}{4})^4

Amount=4800(1+0.02)^4

Amount=4800(1.02)^4

Amount=4800(1.08243216)

Amount=5195.52\ approximately

Hence, enter a payment of 5192.52.

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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