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Aleks04 [339]

Divide $-11 \frac{1}{5}$ by 5. Or multiply it by 1/5.

You get - 56/25 or $-2 \frac{6}{25}$ pounds per month.

5 0

3 years ago

Which weighs more, a pound of feathers or a pound of bricks?

myrzilka [38]

Answer:

They both weigh the same amount.

Step-by-step explanation:

A pound of feathers would land and wouldn't hurt, but a pound of bricks would hurt.

Since they both are exactly one pound, if that is the only question and it's a one-part question then there's your answer.

7 0

3 years ago

If f(x) = 0.8(2 – x), what is the value of f(–2)?

Sergio039 [100]

With functions, you take the number that is in the ( ), so we have f(-2), and take the fomula f(x) = 0.8(2 – x) and everywhere you see an 'x' replace it with the -2 f(–2)=0.8(2 – (-2))

We will have to work with the expression of 0.8(2-(-2) When you want to evaluate these types of expressions, you want to use the Order of Operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. First we have to evaluate the parentheses. What is 2-(-2)
its 4.

ok now you got 0.8 (4)

Now we have the remains of 0.8(4). If a number is within the parenthesis, then it means that we have to multiply the number inside with the number that is outside. What is 0.8*4 its 3.2 soooo your answer is going to be

3.2

5 0

3 years ago

Find the distance from A to C

Vedmedyk [2.9K]

Answer: 11 because the distance between A-B is 8 so because it says it on top, then add 3 to it from B to C

5 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.

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

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

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

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

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

(C)

6 0

2 years ago