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adoni [48]
3 years ago
7

Find the y-intercept for the parabola defined by

Mathematics
1 answer:
olasank [31]3 years ago
6 0

Answer:

(0,6)

Step-by-step explanation:

because a point that lies on y axis has the coordinates of its x = 0.

the therm without the x in the equation of the parabola indicate where it intercepts the y axis

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Simplify 32 ⋅ 35. (4 points)<br><br> 37<br><br> 310<br><br> 97<br><br> 910<br> Pllllsssss help
postnew [5]

Answer:

what the heck is that i rhink you mess up

4 0
3 years ago
Exponent Practice:<br> x^2/3=64
docker41 [41]
X^2/3 = 64
x^{2} =64^{3}
x = 64^{3/2}

it is easier to first find square root of 64 and than power it to 3
square root of 64 is 8 which means our equation now looks like:
x = 8^3
now the answer is after powering 8 to 3:
x = 512

6 0
4 years ago
1 Admission to the local movie theater is $3 for each child and $7 for each adult. A group of 12 people pay $64 admission. How m
Eduardwww [97]

Given:

Admission to the local movie theater is $3 for each child and $7 for each adult.

A group of 12 people pay $64 admission.

To find:

The number of children in a group.

Solution:

Let the number of children be x and number of adults be y.

According to the question,

Total persons : x+y=12          ...(i)

Total cost : 3x+7y=64     ...(ii)

Multiply equation (i) by 7 and subtract the result from (ii).

3x+7y-7(x+y)=64-7(12)

3x+7y-7x-7y=64-84

-4x=-20

Divide both sides by -4.

x=5

Therefore, the number of children is 5.

6 0
3 years ago
Use Algorithm 7 to schedule the largest number of talks in a lecture hall from a proposed set of talks, if the starting and endi
3241004551 [841]

Answer:

If we arrange the talks from the lowest starting time to the highest ending time we get total of 11 talks.

using algorithm 7 we get answer   (1) - (3) - (6) - (9)   the largest number of talks scheduled.

Step-by-step explanation:

arranging the talks from lowest starting time to the highest ending time.

thus,

  1. 9:00 a.m. and 9:45 a.m.
  2. 9:30 a.m. and 10:00 a.m.
  3. 9:50 a.m. and 10:15 a.m.
  4. 10:00 a.m. and 10:30 a.m.
  5. 10:10 a.m. and 10:25 a.m.
  6. 10:30 a.m. and 10:55 a.m.
  7. 10:15 a.m. and 10:45 a.m.
  8. 10:30 a.m. and 11:00 a.m.
  9. 10:45 a.m. and 11:30 a.m.
  10. 10:55 a.m. and 11:25 a.m.
  11. 11:00 a.m. and 11:15 a.m.

we start from the earliest time as

9:00 a.m. and 9:45 a.m which is (1).

After the talk is finished we pick the nearest time for another talk which starts at

9:50 a.m. and 10:15 a.m which is (3).

After this talk we again pick the nearest time for another talk which becomes

10:30 a.m. and 10:55 a.m which is (6).

and lastly

10:45 a.m. and 11:30 a.m which is (9).

Note: we didn't choose other times because we cannot talk at 2 or 3 places at the same time. so we pick another when one talk is finished.

thus the answer is (1) - (3) - (6) - (9) as the largest number of talks scheduled.

4 0
4 years ago
The baseball park is 4200 m from Camden's house. I took him 30 minutes to get there.
Marina86 [1]
The answer is 140 m/min
4 0
3 years ago
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