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Tamiku [17]
3 years ago
11

A³+b³+c³-3abc is = ____​

Mathematics
1 answer:
mezya [45]3 years ago
7 0

Answer:

open the photo u will find answer

okkk...

plz mark as brainliest plz friend

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What is the true solution to the equation below?
Marina CMI [18]

Answer:

The solution is:

  • x=4

Step-by-step explanation:

Considering the expression

lne^{lnx}+lne^{lnx}^{^2}=2ln8

\ln \left(e^{\ln \left(x\right)}\right)+\ln \left(e^{\ln \left(x\right)\cdot \:2}\right)=2\ln \left(8\right)

\mathrm{Apply\:log\:rule}:\quad \:log_a\left(a^b\right)=b

\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2

\ln \left(x\right)+\ln \left(x\right)\cdot \:2=2\ln \left(8\right)

\mathrm{Add\:similar\:elements:}\:\ln \left(x\right)+2\ln \left(x\right)=3\ln \left(x\right)

3\ln \left(x\right)=2\ln \left(8\right)

\mathrm{Divide\:both\:sides\:by\:}3

\frac{3\ln \left(x\right)}{3}=\frac{2\ln \left(8\right)}{3}

\ln \left(x\right)=\frac{2\ln \left(8\right)}{3}.....A

Solving the right side of the equation A.

\frac{2\ln \left(8\right)}{3}

As

\ln \left(8\right):\quad 3\ln \left(2\right)

Because

\ln \left(8\right)

\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3

⇒ \ln \left(2^3\right)

\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

\ln \left(2^3\right)=3\ln \left(2\right)

So

\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}

\mathrm{Multiply\:the\:numbers:}\:2\cdot \:3=6

          =\frac{6\ln \left(2\right)}{3}

\mathrm{Divide\:the\:numbers:}\:\frac{6}{3}=2

          =2\ln \left(2\right)

So, equation A becomes

\ln \left(x\right)=2\ln \left(2\right)

\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)

         =\ln \left(2^2\right)

         =\ln \left(4\right)

\ln \left(x\right)=\ln \left(4\right)

\mathrm{Apply\:log\:rule:\:\:If}\:\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\:\mathrm{then}\:f\left(x\right)=g\left(x\right)          

x=4

Therefore, the solution is

  • x=4
6 0
3 years ago
Read 2 more answers
Which equation is in standard form?
evablogger [386]
The answer is C. 2x+3y=12
3 0
3 years ago
From a survey of 60 students attending a university, it was found that 9 were living off campus, 36 were undergraduates, and 3 w
lana66690 [7]
I think the question has to do with the number of students who are attending the university but is neither an undergraduate nor living off-campus. To help us solve this problem, we use the Venn diagram as shown in the picture. The intersection of the 2 circles would be 3 students. The students in the 'students living off-campus' circle would be 9 - 2 = 6, while the undergraduate students would be 36-3 = 33. The total number of students inside all the circles and outside the circles should sum up to 60 students.

6 + 3 + 33 + x = 60
x = 60 - 6 - 3 - 3
x = 18 students

Therefore, there are 18 students who are neither an undergraduate nor living off-campus

8 0
3 years ago
How do I solve 2x+3y=7, x-y=1
Fittoniya [83]
<span>2x+3y=7
x-y=1

</span>2x+3y=7
x=1+y

2*(1+y)+3y=7
x=1+y

2+2y+3y=7
x=1+y

5y=5       |(:5)
x=1+y

y=1
x=2

CHECK:
2x+3y=7
x-y=1

2*2 + 3*1 = 7     

x-y = 1
2-1 = 1

5 0
3 years ago
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Which expression is equivalent to(2^3)^-5
Nady [450]

Answer:

hope this helps if not im sorry

Step-by-step explanation:

(2^3)^5

=2^(3*5)

= 2^15

3 0
3 years ago
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