All categories

3 years ago

Please help me on these using conversion factors. And please include work and tell me which one is greater

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

8 0

Answer: 2m > 6ft, 10kg > 10lb

Step-by-step explanation:

6 ft is 72 inches (12 inches per foot)

2 m is a little over 78 inches (39+ inches per meter)

10kg is 22lb (2.2 lb per kg)

You might be interested in

A tractor was purchased for $30,000 six years ago. what is the value of the tractor today if the depreciation rate was 2.5%?

MrRissso [65]

Initial cost = $30,000
Depreciation rate = 2.5%

Depreciation expense per year = 30,000*2.5/100 = $750
In six years,
Depreciation = 6*750 = $4,500

Value of the tractor = Initial cost - Depreciation = $30,000 - $4,500 = $25,500

6 0

3 years ago

Sam traveled a distance of 3724 feet,how many miles did she travel if 1 mile =5,280 feet?

lilavasa [31]

Answer:

0.705303

Step-by-step explanation:

divide 3724 by 5280

8 0

3 years ago

4y-28 pls help find y value

mina [271]

Answer:

4y-28 = 0

4y = 28

y=28/4

y = 7

5 0

2 years ago

Read 2 more answers

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.

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