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3 years ago

PLEASE HELP!!! DONT IGNORE ME!!!

Mathematics

1 answer:

MrRa [10]3 years ago

5 0

1. 2x^2-x-15=0
A = 2
B = -1
C = -15

2. 10^2-2x=0
A = 10
B = -2
C = 0

3. x^2–3x-40=0
A = 1
B = -3
C = -40

4. x^2+3x-2=0
A = 1
B = 3
C = -2

5. 4x^2-17x+8=0
A = 4
B = -17
C = 8

Remember the equations should look like Ax^2+Bx+C=0 (in this order!!)

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anyanavicka [17]

Answer:

x is 5 and y is 6

Step-by-step explanation

we find 2/3 factors that equal 10. 2 and 3. the only other number that can be added that equals 10 is 5. i apply the same logic to y

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4 years ago

The third graders collected 754 cans the fourth graders collected 592 cans estimate how many more cans the third graders collect

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There are 22 4th grade students and 26 fifth grade students who became cheerleaders each cheer leading group will have 8 members

BlackZzzverrR [31]

Answer:

The equation to find the number of groups can be formed is given as: $n=\frac{22+26}{8}$

where $n$ represent the number of groups that can be formed.

We get $n=6$ on solving

Step-by-step explanation:

Number of 4th grade students who become cheerleader = 22

Number of 5th grade students who become cheerleader = 26

Number of cheerleaders in each cheer leading groups = 8

We need to divide the total number of cheer leaders into groups of 8.

Let $n$ represent the number of groups that can be formed.

Total number of students who want to become cheerleaders = $22+26$

So, equation to find the number of groups can be given as :

$n=\frac{22+26}{8}$

⇒ $n=\frac{48}{8}$

⇒ $n=6$

Thus,a total of 6 cheer leading groups can be formed.

3 0

3 years ago

The amount of energy, E, required to raise the temperature, T, of a container of water is E=240T, where E is measured in calorie

SSSSS [86.1K]

E=240T

For the lower limit of energy:
E=12,000 cal

so 12000 = 240 T >>> T = 50 degrees Celsius

For the upper limit of energy:
E=18,000 cal

so 18000 = 240 T >>> T = 75 degrees Celsius

The difference in temperature = 75 - 50 = 25 degrees Celsius

I hope I helped you.

5 0

4 years ago