Answer:
D) 3x^2 - 12
Step-by-step explanation:
Using PEMDAS;
There is no need to evaluate the part of the equation (x^2 - 8) because is no need to, as it is already in its simplest form.
We must evaluate the part of the equation continuing with, "- (-2x^2+4)," as it is not in its simplest form.
Evaluating "- (-2x^2+4)":
Step 1: Distributing the negative
Once distributing the negative symbol amongst the values within the parenthesis according to PEMDAS, we get "2x^2 - 4" as the product.
Step 2: Consider the rest of the equation to evaluate
Since the part of the equation is still in play here as it is a part of the original equation to be solved, we must evaluate it as a whole to get the final answer.
Thus,
x^2 -8 + 2x^2 - 4 = ___
*we can remove the parenthesis as it has no purpose, since it makes no difference.
Evaluating for the answer, we get,
x^2+2x^2 + (-8 - 4) = 3x^2 - 12
Hence, the answer is D) 3x^2 - 12.
Answer:
2 meters
Step-by-step explanation:
We need to use trigonometry for this. The appropriate one would be tangent, which is (opposite side) divided by (adjacent side).
In this case, the opposite side of angle A is BC, which is 6 meters. The adjacent side of angle A is AB, which is the ground. Since we don't know its length, we call it x.
Now, we write:
= 6/x
To solve, we just multiply both sides by tan(x) and x:
x = 6/[tan(72)] ≈ 1.95 meters ≈ 2 meters.
Thus, the answer is 2 meters.
Hope this helps!
Answer:
Step-by-step explanation:
Given:
Area = 3x^3 - 16x^2 + 31x - 20
Base:
x^3 - 5x
Area of trapezoid, S = 1/2 × (A + B) × h
Using long division,
(2 × (3x^3 - 16x^2 + 31x - 20))/x^3 - 5x
= (6x^3 - 32x^2 + 62x - 40))/x^3 - 5x = 6 - (32x^2 - 92x + 40)/x^3 - 5x = 2S/Bh - Ah/Bh
= 2S/Bh - A/B
= (2S/B × 1/h) - A/B
Since, x^3 - 5x = B
Comparing the above,
A = 32x^2 - 92x + 40
2S/B = 6
Therefore, h = 1
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
T-m=25
3m=J
275=t+j+m
Matt spent 50 dollars
Jeff spent 150
and Tim spent 75
