3 years ago

At the end of a 15-year term, what is the balance of a $3,000 investment at a simple interest rate of 1.5 percent?

1 answer:

5 0

Find how much interest is earned for one year: $3000 x .015 = $45
Use that answer to find how much interest is earned for 15 years: $45 x 15 years = $675
Add original investment to interest for balance: $3000 + $675 = $3675

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What is the solution to the system of equations? Use the substitution method. {y=−2x+5 3x−4y=2

GaryK [48]

We have that
y=−2x+5 --------> equation 1
3x−4y=2---------> equation 2
substituting 1 in 2
3x-4*(-2x+5)=2--------> 3x+8x-20=2--------> 11x=22-----> x=2
y=-2x+5-----> -2*2+5-----> y=1

the solution is
x=2
y=1

5 0

3 years ago

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Elden [556K]

Answer:

Step-by-step explanation:

The correct answer is y = 3x, where y is the total cost and x is the number of pounds of oranges purchased.

5 0

2 years ago

How do you simplify: <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%28125%20%5E%7B2%7D%20%7D%20%29%20%20%20%5E%7B%20-%20

Nady [450]

Answer:

$\huge\boxed{\sqrt{(125^2)^{-\frac{1}{3}}}=\dfrac{1}{5}}$

Step-by-step explanation:

$\sqrt{(125^2)^{-\frac{1}{3}}}\qquad\text{use}\ (a^n)^m=(a^m)^n\\\\=\sqrt{\left(125^{-\frac{1}{3}\right)^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=125^{-\frac{1}{3}}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{1}{125^\frac{1}{3}}\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\dfrac{1}{\sqrt[3]{125}}=\dfrac{1}{5}\qquad\text{because}\ 5^3=125$