All categories

3 years ago

Solve open parentheses square root of 6 close parentheses to the 8 x power = 216x−3.

Mathematics

2 answers:

Inessa05 [86]3 years ago

6 0

Answer:

-9

Step-by-step explanation:

ololo11 [35]3 years ago

4 0

We can make the base of the equation same on both the sides and then equate the exponents to get the answer as shown below:

$(\sqrt{6})^{8x}=(216)^{x-3} \\ \\ (\sqrt{6})^{8x}=(6^{3})^{x-3} \\ \\ (6^{\frac{1}{2}}) ^{8x}=(6)^{3x-9} \\ \\ (6)^{4x} =(6)^{3x-9} \\ \\ 4x=3x-9\\ \\ 4x-3x=-9\\ \\ x=-9$

Therefore, first option gives the correct answer

You might be interested in

Will give, Branliest.<br><br>Thanks, <br>:)

Nikolay [14]

Answer:

250p-195

Step-by-step explanation:

did the quiz

8 0

3 years ago

Read 2 more answers

Jenna earned an 80% on her Benchmark Test. If she answered 36 questions correctly, how many questions were on the test?

spin [16.1K]

There were 45 questions on the test. Hope this helps! <3

3 0

3 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago

Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If thi

defon

7978 years

Step-by-step explanation:

A = A02^(-t/hl)

where hl = half-life.

Dividing both sides by A0 and taking the logarithm, we get

ln(A/A0) = -(t/hl)ln2

or solving for t,

t= -(hl)ln(A/A0)/ln2

note that A/A0 = 0.38

t = -(5715 yrs)[ln(0.38)/ln2]

= 7978 years

7 0

3 years ago

Please help, The first one to answers gets 10 points

babymother [125]

Answer:

6.75-2.33-1.75=2.67 hours

Step-by-step explanation:

5 0

3 years ago