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4 years ago

Could someone help me

Mathematics

1 answer:

frutty [35]4 years ago

8 0

Answer: um need the question to answer it kk

Step-by-step explanation:

You might be interested in

what is the probabilty that the sum of the two rolls is a prime number, given that atleast one of the rolls is a 3?

Sav [38]

the answer is you would have a 1/3 chance

3 0

3 years ago

Shayna's school is selling tickets to a fall musical. on the first day of tickets sales the school sold 14 adult tickets and 1 s

KengaRu [80]

Answer:

The answer is: Each adult ticket is $7 and each student ticket is $9.

Step-by-step explanation:

Let a = the number of adult tickets. Let s = the number of student tickets.

For day 1:

14a + 1s = $107

Solve for s:

s = 107 - 14a

For day 2:

12a + 5s = $129

By substitution:

12a + 5(107 - 14a) = 129

12a + 535 - 70a = 129

-58a = 129 - 535

-58a = -406

a = -406 / -58 = $7 for adult tickets

Solve for s:

s = 107 - 14a

s = 107 - 14(7)

s = 107 - 98

s = $9 for student tickets

Proofs:

For day 1:

14a + 1s = $107

14(7) + 1(9) = 107

98 + 9 = 107

107 = 107

For day 2:

12a + 5s = $129

12(7) + 5(9) = 129

84 + 45 = 129

129 = 129

3 0

3 years ago

If the quadratic formula is used to find the solution set of 3x2 + 4x - 2 = 0, what are the solutions?

7nadin3 [17]

Answer:

$x=\frac{-2}{3} \pm \frac{\sqrt{10}}{3}$

Step-by-step explanation:

Compare $ax^2+bx+c$ to $3x^2+4x-2$ .

We have $a=3,b=4,c=-2$ .

The quadratic formula is for solving equations of the form $ax^2+bx+c=0$ and is $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ .

So we are going to plug in our values in that formula to find our solutions,x.

If you want to notice it in parts you can.

Example I might break it into these parts and then put it in:

Part 1: Evaluate $b^2-4ac$

Part 2: Evaluate $-b$

Part 3: Evaluate $2a$

------Let's do these parts.

Part 1: $b^2-4ac=(4)^2-4(3)(-2)=16-12(-2)=16+24=40$ .

This part 1 is important in determining the kinds of solutions you have. It is called the discriminant. If it is positive, you have two real solutions. If it is negative, you have no real solutions (both of the solutions are complex). If it is 0, you have one real solution.

Part 2: $-b=-4$ since $b=4$ .