Convert an annual rate of 0.224% to a generation rate (20 years)

one leg of a right triangle is 5 millimeters shorter than the longer leg and the hypotneuse is 5 millimeters longer than the lon

Makovka662 [10]

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.
-----------------------------
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

7 0

2 years ago

Which expression demonstrates the use of the commutative property of addition in the first step of simplifying the expression (–

madam [21]

The first step of simplifying the expression would be summing up the "i"s. They are separate ("+i" and "+5i") but you can still sum them up though they are apart. This is thanks to the commutative property of addition which allows summing the terms of an expression with several additions in whatever order it is prefered. Commutative means that the terms can be interchangeable.

4 0

3 years ago

If P(x) = P<img src="https://tex.z-dn.net/?f=_%7Bn%7D" id="TexFormula1" title="_{n}" alt="_{n}" align="absmiddle" class="latex-f

seropon [69]

If $P(x) = p_nx^n + p_{n-1}x^{n-1}+\ldots+p_0$ is divided by $(x-a)$ , then $P(x) = (x-a) \cdot Q(x) + R(x)$ for some polynomials $Q,R$ . Moreover, $\deg R < 1$ (because $\deg (x-a) = 1$ ), so there exists $\alpha \in \mathbb{R}$ such that $R(x) = \alpha$ for all $x \in \mathbb{R}$ . But if we calculate $P(a)$ , it turns out that $P(a) = (a-a)\cdot Q(a) + \alpha$ , so $R(x) = \alpha = P(a)$ . $\blacksquare$