Which is a solution to the following system of equations?

Mathematics

1 answer:

disa [49]3 years ago

3 0

Answer:

I think number B

Step-by-step explanation:

I plugged it in the calculator. And then they both were zero.

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Solving an Equation for a Rate Problem Stu hiked a trail at an average rate of 3 miles per hour. He ran back on the same trail a

ohaa [14]

Given:
speed = distance / time

walking: 3 miles per hour
running 5 miles per hour
for a total of 3 hours.

3t = 5(3-t)
3t = 15 - 5t
3t + 5t = 15
8t = 15
t = 15/8
t = 1 7/8

7/8 * 60 mins = 52.50 minutes

t = 1 hour and 52.5 minutes

It took Stu 1 hour and 52.5 minutes to hike the trail one way.

6 0

3 years ago

Read 2 more answers

NEED HELP PLZ HELP ME ….

Katyanochek1 [597]

Answer:

R= -0.95

Step-by-step explanation:

HOPE THIS HELPS :)

7 0

2 years ago

Another question I'm stuck on

qwelly [4]

M=(y2-y1)/(x2-x1) m=(-4-4)/(3-1) m=-8/2 m=-4 so the answer should be first one

7 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$