Answer: (3b + 5)(3b - 5)
Because both terms, 9b² and 25, are perfect squares, you can factor by taking the square roots of both terms.
The square root of 9b² is 3b (3b × 3b = 9b²).
The square root of 25 is 5 (5 × 5 = 25).
9b² - 25 has a negative, so the factored expression would be
(3b + 5)(3b - 5). The signs (+ and -) alternate in this case because the expression, 9b² - 25, has no middle term.
You can check your work by using FOIL. See the attachment below.
F irst
O utside
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Given A = {a, e, i, o, u} and B = {a, l, g, e, b, r}, find A ∪ B.
harkovskaia [24]
Ahh..this is sets topics - A U B = all the elements found in A and B. But do note, do not repeat the elements if it is the same. And if the question were to ask : n(AUB) = total number of elements found in A and B.
Well, the best way to solve this would involve the Pythagorean theorem.
First, take the length of the hypotenuse (the string in this case) and square it. Do the same to the side length we are given (the kites height above the ground).
We then subtract the squared height from the squared hypotenuse. We should be left with 1,781m.
All we have to do now is find the square root of 1,781m.
Your answer for b is 42.20189569m.
Given:
Bees scored 9 less than three times as many points as the Hornets.
Wasps scored 28 more points than the Hornets.
Together the three teams scored 184 points.
To find:
The required equation for this scenario.
Step-by-step explanation:
Let x represent the number of points scored by the Hornets.
Bees scored 9 less than three times as many points as the Hornets.
Bees score = 3x-9
Wasps scored 28 more points than the Hornets.
Wasps score = x+28
Together the three teams scored 184 points.
Therefore, the required equation is .
9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
