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.
160 cm because if I am correct, then two of the sides are labeled as 10 cm. Which means that they are both the same measurement.
Answer: where is the figure tot he right???
Step-by-step explanation:
Answer:
307 paces
Step-by-step explanation:
First, lets convert Robert's pace to metres. (cm > m)
75cm = (75/100)m
= 0.75m
Number of Full Paces = Length of Bridge / 1 Full Pace
=
See , we must have full paces, and in this case we are finding the minimum number of paces, so we will be looking at the lowest possible whole number, which in this case, is 307 paces.
Step-by-step explanation:
Not entirely sure, but I believe something like this:
∠1 + ∠2 = 180° since they are a linear pair;
Since ∠1 = ∠2, we can say:
∠1 + ∠1 = 180°
2(∠1) = 180°
∠1 = 90°
And since:
∠1 = ∠2
∠2 = 90°
So the angle between t and s is 90°;
An angle of 90° means t and s are perpendicular (⊥)
