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Tcecarenko [31]
3 years ago
5

Sam and Bruno were computing how many kilometers they rode in the 3 bike trips they took last month. In order, they rode 45.7, 4

0.9, and 38 miles. Their solutions are shown.
Mathematics
1 answer:
Eddi Din [679]3 years ago
4 0

Answer:

We are given number of miles the 3 bike trips they took last month = 45.7, 40.9, and 38 miles.

Sam and Bruno were computing how many kilometers.

Also one kilometer equals 0.621 miles.

From this we can see than 1 kilometer is smaller unit and a mile is a larger unit.

In order to convert number of miles into kilometers, we need to divide each number of mile by 0.621 miles.

We always divide a larger value by a smaller value to get the smaller unit.

Therefore,

Bruno is correct. When going from a larger unit to a smaller unit you need to divide.

Step-by-step explanation:

Hopefully this helped, if not HMU and I will get u a better answer.

<em>-Have a great day! :)</em>

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Alex scored 70% in a spelling test with 20 quetions
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Answer:

This means that Alex answered 14 questions correctly

Step-by-step explanation:

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3 years ago
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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

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
Help me solve this problem
eduard

Answer:

b.   (-1, 5)

Step-by-step explanation:

Since the first equation is already solved for y, we can use substitution.

Substitute y in the second equation with 3x + 8.

5x + 2y = 5

5x + 2(3x + 8) = 5

5x + 6x + 16 = 5

11x = -11

x = -11/11

x = -1

Now substitute x in the first original equation by -1 and solve for y.

y = 3x + 8

y = 3(-1) + 8

y = -3 + 8

y = 5

Answer: (-1, 5)

8 0
11 months ago
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