Jake bought a mountain bike on sale. He saved $72 on the bike,

An item is regularly priced at 85$. It is on sale for 80% off the regular price. How much is discounted from the regular price?

laila [671]

$68 is discounted from the regular price.

To find this, you first divide 80% by 100 to get 0.8 (the decimal form of the percentage). Then, simply multiply 0.8 by 85 (or vise verse) to get $68. Since the question is asking for how much is discounted from the regular price, the final answer would be $68.

I hope this helps!

4 0

2 years ago

SARAH MAKES 7 MEIKLES TO RAISE MONEY FOR A CHARITY.SHE CAN MAKE 7 MEIKLES IN a DAY AND ALREADY HAS 4 MEIKLES MADE.

OlgaM077 [116]

It would be 130/7+4 because 130 is the total divide it by how many she does and add how many she already has

6 0

2 years ago

Plz plz help I’m failing math and need help

diamong [38]

Answer:

5 sweat shirts- $100

8 sweat shirts- $160

9 sweat shirts- $180

12 sweat shirts- $240

Step-by-step explanation:

Each sweat shirt costs $20

5 0

2 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.

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