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Orlov [11]
3 years ago
14

There are 5 designs of necklaces available at a jewelry store. Each design is available in 3 types of stones. How many different

combinations of 1 design and 1 stone of necklace can you have?
Mathematics
1 answer:
JulijaS [17]3 years ago
3 0

Answer:

3 stone necklaces

Step-by-step explanation:

so theres 5 designs and 3 stones you can only have 1 stone with one if your looking to put one on each necklace there will be three with one stone and two without

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Please help! It will be appreciated!
AysviL [449]
It’s d, looks like you got it already! With ratios, just pay attention to not only the numbers but order of the words as well, so you can be sure the ratio follows the problem.
8 0
3 years ago
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One T-shirt is sold for every 5 people who attend a concert.
docker41 [41]
5:1 people to T Shirt
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Work out the value of 5x-2y when x=-2 and y=-3
max2010maxim [7]
I guess this how it would go
<span><span><span>(5)</span><span>(<span>−2</span>)</span></span>−<span><span>(2)</span><span>(<span>−3</span>)</span></span></span>

So your answer should be <span>−4</span>

3 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
If k=100 - 4n what’s the value when n=60
dexar [7]
To solve this problem, start by plugging in your given n value.
This results in:
k = 100 - 4(60)

Next, simplify the parentheses by multiplying the -4 and the 60 together.
This leaves you with:
k = 100 - 240

To simplify for k, subtract 240 from 100 to get:
k = -140

So your final answer is k = -140 when n = 60
4 0
3 years ago
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