Answer:
201
Step-by-step explanation:
3.14*8*8=201
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Answers:
- C) x = plus/minus 11
- B) No real solutions
- C) Two solutions
- A) One solution
- The value <u> 18 </u> goes in the first blank. The value <u> 17 </u> goes in the second blank.
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Explanations:
- Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
- There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
- The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
- We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
- Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
Answer:
The side s has a length of 4 and side q has a length of 4
⇒ F
Step-by-step explanation:
In the 30°-60°-90° triangle, there is a ratio between its sides
side opp (30°) : side opp (60°) : hypotenuse
1 : : 2
In the given triangle
∵ The side opposite to 30° is s
∵ The side opposite to 60° is q
∵ The hypotenuse is 8
→ Use the ratio above to find the lengths of s and q
side opp (30°) : side opp (60°) : hypotenuse
1 : : 2
s : q : 8
→ By using cross multiplication
∵ s × 2 = 1 × 8
∴ 2s = 8
→ Divide both sides by 2
∴ s = 4
∴ The length of s is 4
∵ q × 2 = × 8
∴ 2q = 8
→ Divide both sides by 2
∴ q = 4
∴ The length of q is 4
a) 
because it is equal to the area of the shaded region between X=4 and X=6, and the probability that X falls within some interval is given by the area under the PDF.
b) 
because the shaded region is a rectangle of height 1/5 (by virtue of X following a uniform distribution over the interval [2, 7], which has length 5).
