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lora16 [44]
3 years ago
7

Express the following relationship as a rate.

Mathematics
1 answer:
defon3 years ago
3 0
It’s $1.15 per attempt
You might be interested in
6v+13b=377<br> 4v+9b=259
aleksandr82 [10.1K]

Answer:

v = 13

b = 23

Step-by-step explanation:

To solve this system of equations, we can negate one equation and add it to other to cancel out one variable.

Looking at v and b, the coefficients are 6, 4 and 13, 9, respectively.

It would be easier to use the LCM of 6 and 4

E_1 : 6v + 13b = 377\\E_2:  4v + 9b = 259\\\\4E_1 + (- 6)E_2:\\24v + 52b = 1508\\-24v - 54b=-1554\\\\0v -2b = -46\\-2b = -46\\b = 23\\\\4v + 9(23) = 259\\4v + 207 = 259\\4v = 52\\v = 13\\\\6(13) + 13(23)\\78 + 299\\377

6 0
3 years ago
What is the least common multiple of 9 , 12, 15
True [87]

Answer:

3

Step-by-step explanation:

9÷3 , 12÷3 , 15÷3

= 3

done

8 0
3 years ago
Before sending track and field athletes to the Olympics, the U.S. holds a qualifying meet.
Leto [7]

Answer:

A

Step-by-step explanation:

I took the test just for you

8 0
3 years ago
Read 2 more answers
Joe borrowed $8,000 from the bank at a rate of 7% simple interest per year. How much interest did he pay in 9 years?
AleksAgata [21]

Answer:

5,040.00

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 7%/100 = 0.07 per year.

Solving our equation:

A = 8000(1 + (0.07 × 9)) = 13040

A = $13,040.00

The total amount accrued, principal plus interest, from simple interest on a principal of $8,000.00 at a rate of 7% per year for 9 years is $13,040.00.

3 0
3 years ago
If a right triangle's hypotenuse is 17 units long, and one of its legs is 15 units long, how long is the other leg?
Vesna [10]

Answer:

8 units

Step-by-step explanation:

Hello!

So, there's a formula we can apply to right-angled triangles: Pythagorean's theorem. It states that  c = \sqrt{{a}^2 + b^{2} }, where <em>c</em> is the hypotenuse and <em>a</em> and <em>b </em>are the legs of the triangle.

So, from the problem,  if <em>c </em>= 17 and <em>a </em> = 15, then, we're solving for <em>b</em>. So we'll rewrite the theorem to solve for <em>b</em>.

{c}^2 = {a}^2+{b}^2\\{c}^2-{a}^2={b}^2\\{b} = \sqrt{{c}^2-{a}^2}

Okay, so now we have isolated the theorem for <em>b. Let's </em>plug in our values for <em>c </em>and <em>a</em>.

b = \sqrt{{17}^2-{15}^2}\\b = \sqrt{289-225}\\b = \sqrt{64}\\b = 8

So, using the theorem, we found <em>b</em> = 8. To check our work, let's plug in <em>b</em> and <em>a</em> and solve for <em>c.</em>

<em />c = \sqrt{{a}^2+{b}^2}}\\c = \sqrt{{15}^2+{8}^2}\\c = \sqrt{225+64}\\c = \sqrt{289}\\c = 17\\<em />

So, we got our hypotenuse to equal 17 units, which is correct! So, our <em>b</em> is correct too. Awesome

7 0
3 years ago
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