Need help with a few Algebra problems because I can't really do this atm. Problem 1:
x^2 - 13x - 41 = 4
Step 1: x^2 - 13x - 36 = 0
Step 2: (x-9)(x+4) = 0
Step 3: x - 9 = 0 x + 4 = 0
Step 4: x = 9 x = 4

There is a mistake in there. How should it be fixed?

Problem 2:
x^3 - 16x = 0
Step 1: (x-4)(x+4)
Step 2: x-4 = 0 x+ 4 = 0
Step 3: x = 4 / x = -4

If one of these steps are incorrect, list it and tell me how to fix it.

Problem 3:
2x^2 - 7x 3 = 1
Step 1: 2x^2 - 7x - 4 = 0
Step 2: x^2 - 7x - 4 = 0
Step 3: (x-8)(x+1) = 0
Step 4: (x-)(x+) = 0
Step 5: (x-2)(4x+1) = 0
Step 6: x - 2 = 0 4x + 1 = 0
Step 7: x = 2 x = -

One of these steps is incorrect, Tell me how to fix it along which step it is.

Show work for all 3 problems.

1

x = 1

step-by-step explanation:

0 ÷ x = 0

0 ÷ 1 = 0

a) at most means it can cost $12 or less

so the inequality sign would be ≤

you would then need to multiply the number of toppings by the 0.35 for each one and add it to the cost of the cheese pizza.

the equation would be 9.99 + 0.35x ≤ 12

b) using the above equation:

9.99 + 0.35x ≤ 12

subtract 9.99 from each side:

0.35x ≤2.01

divide both sides by 0.35:

x ≤ 2.01 / 0.35

x ≤5.74

at most you could have 5 toppings.

the 5 topping pizza would cost: 0.35 x 5 = 1.75 + 9.99 = $11.74

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen