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3 years ago

6

Which of the following is true regarding the solution to the logarithmic equation below?

2 answers:

Gemiola [76]3 years ago

8 0

Although the options are not provided, the true statement is:
x = 5 is a true solution because log₂(16) = 4
In order to check if a solution is true, we may plug it back into the original equation. When we plug x into the equation we get:
log₂(5+11) = 4
log₂(16) = 4
4 = 4

As the left hand side of the equation is equal to the right hand side, the solution is true.

Varvara68 [4.7K]3 years ago

8 0

Answer:

c on edge

Step-by-step explanation:

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2 years ago

Can you guys please help me. i don’t need the explanation i just need the answer

Fittoniya [83]

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36

Step-by-step explanation:

AOB + BOC = AOC

AOC = 90 since the lines are perpendicular

6x-12 + 3x+30 = 90

Combine like terms

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5 0

3 years ago

Read 2 more answers

- 3/5 + - 3/4 can somebody help plz

Gnesinka [82]

<span>- 3/5 + - 3/4
= (-12/20) = (-15/20)
= - 27/20
= - 1 7/20</span>

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3 years ago

Read 2 more answers

WHO CAN solve it Please !

Mariulka [41]

Answer:

a) True

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha$ <em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em> cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha$ <em></em>

<em> = </em> $\int\limits^\pi _0 {cos\alpha } \, dx$ <em></em>

<em>Now, Integrating </em>

<em> </em> $= ( sin\alpha )_{0} ^{\pi }$ <em></em>

<em> = sin π - sin 0</em>

<em> = 0-0</em>

<em> = 0</em>

<u><em>Final answer:-</em></u>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

<em></em>

<em></em>

5 0

2 years ago