A example can be Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $25 each week for food, clothes, and movie tickets. The answer would be Any more than 12 weeks and his account balance would be less than $200. Any number of weeks less than 12 and his account would stay above $200.
Answer:
7.11
-7.1
Step-by-step explanation:
Diameter = 7 cm
radius = 7÷2 = 3.5
Volume of cone = 1/3
Volume of cone = 1/3
= 115 cubic inches (Answer B)
The answer, in short, is that the short leg equals 15 mm, the long leg equals 20 mm, and the hypotenuse equals 25mm. but if you'd like to see how I solved it, here are the steps.
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The Pythagorean theorem (also known as Pythagoras's Theorem) can be used to solve this. This theorem states that one leg or a right triangle squared plus the other side of that same triangle squared equals the hypotenuse of that triangle squared. To put it in equation form, L² + L² = H².
Let's call the longer leg B, the shorter leg A, and the hypotenuse H.
From the question, we know that A = B - 5, and H = B + 5.
So if we put those values into an equation, we have (B - 5)² + B² = (B + 5)²
Now, to solve. Let's square the two terms in parentheses first:
(B² - 5B - 5B + 25) + B² = B² + 5B + 5B + 25
Now combine like terms:
2B² -10B + 25 = B² + 10B + 25
And now we simplify. Subtract 25 from each side:
2B² - 10B = B² + 10B
Subtract B² from each side:
B² - 10B = 10B
Add 10B to each side:
B² = 20B
And finally, divide each side by B:
B = 20
So that's the length of B. Now to find out A and H.
A = B - 5, so A = 15.
H = B + 5, so H = 25.
And your final answer is A = 15mm, B = 20mm, and H = 25mm
Depending on the values of 'r', 't', and 'e', the numerical value of that expression
might have many factors.
For example, if it happens that r=5, t=1, and e=4 for an instant, then, just
for a moment, (r + t)(e) = (5+1)(4) = 24, and the factors of (r+t)(e) are
1, 2, 3, 4, 6, 8, 12, and 24 . But that's only a temporary situation.
The only factors of (r+t)(e) that don't depend on the values of 'r', 't', or 'e' ,
and are always good, are (<em>r + t</em>) and (<em>e</em>) .
