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frosja888 [35]
2 years ago
13

;];];];];];;];;]] math

Mathematics
1 answer:
SOVA2 [1]2 years ago
6 0

Answer:

is epic

Step-by-step explanation:

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Giving brainly to correct answer
Elena L [17]

Answer:

The answer is the second one

im sorry if this is wrong

Step-by-step explanation:

4 0
2 years ago
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Maria had a length of fabric to make banners.
kirill115 [55]

Answer:

the fabric was 16.5 m long

Step-by-step explanation:

2.75 * 6 = 16.5

ANSWER CHECK:

16.5 / 6 = 2.75

I hope this helps :}

Brainliest is very much appreciated thank u fran

5 0
3 years ago
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Answer 10,11,12,13,14
QveST [7]
10
13.50 per set
+.45 per day value
she got 20 sets
5 days after would be
13.50 ×20=270
value of 20 sets at 13.50 a set on day one was 270.00
.45×20=9.00
the 20 sets gain 9 dollars a day
9×5=45
9 dollars a day times 5 days is 45 dollars
45+270=315

11
4 0
3 years ago
HELP Combining like terms solve all please :) c
In-s [12.5K]

Answer:

11. x = -16

12. k = 6

13. x = -19

14. x = -6

15. x = -20

16. Combining like terms isn't to be used on this type of problem. I'm sorry, can you guess on this one?

17. x = 19

18. n = -10

19. b = 11

20. n = 4

21. r = -6

22. n = -4

Again super sorry about question 16 :(

6 0
2 years ago
Read 2 more answers
A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3
fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7

8 0
2 years ago
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