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wel
3 years ago
11

Write Five hundred thirty-eight thousandths in numbers​

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0
.538 should be correct
You might be interested in
Hello please help! posted picture of question
notsponge [240]
The answer to this question is FALSE.

Domain is the set of all the numbers that we can input to the function or that can be used in place of x. The numbers which make the function undefined are excluded from the domain.

In the given exponential function, there is no any value of x which will make the function undefined, so the domain of the function if set of All real numbers. In general, domain of exponential functions is Set of All real numbers. 
5 0
3 years ago
Find the exact value of tan 15 degrees
pentagon [3]
<span>tan(15) = 
sin(15) / cos(15) = 
sin(45 - 30) / cos(45 - 30) = 
[ sin(45)cos(30) - sin(30)cos(45) ] / [ cos(45)cos(30) + sin(45)sin(30)] 

Since sin(45) = cos(45) = √2/2, you can just factor that out from the top and bottom 
[ cos(30) - sin(30) ] / [ cos(30) + sin(30)] 
[ √3/2 - 1/2 ] / [ √3/2 + 1/2] 
(√3 - 1) / (√3 + 1) 
(√3 - 1)^2 / (√3+1)(√3 - 1) 
(√3 - 1)^2 / (3 - 1) 
(3 - 2√3 +1) / 2 
2 - √3 

There's also a formula for tan(a-b), but I couldn't remember it off hand.</span>
4 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
Is there anybody taking the algebra test in texas i would really appreciate the answeres​
s344n2d4d5 [400]

Answer:

which one, i just took a test, and im from texas

Step-by-step explanation:

6 0
2 years ago
Please help fast will give 5 stars and thank you
aniked [119]

Answer:

(1,-1)

Step-by-step explanation:

Hope it helps

7 0
3 years ago
Read 2 more answers
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