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fredd [130]
3 years ago
14

Calculus= Integrate 5^x dx please break down how answer is 1/6x^6+C

Mathematics
1 answer:
aleksley [76]3 years ago
4 0

Answer:

\int\limits {5^x} \, dx = \frac{5^x}{ln\ x} + c

Step-by-step explanation:

Note that the integral of 5^x is not \frac{1}{6}x^6 + c

The solution is as follows:

Given

5^x

Required

Integrate

Represent the given expression using integral notation

\int\limits {5^x} \, dx

This question can't be solved directly;

We'll make use of exponential rules which states;

\int\limits {a^x} \, dx = \frac{a^x}{ln\ x} + c

By comparing \int\limits {5^x} \, dx with \int\limits {a^x} \, dx;

we can substitute 5 for a;

Hence, the expression \int\limits {a^x} \, dx = \frac{a^x}{ln\ x} + c becomes

\int\limits {5^x} \, dx = \frac{5^x}{ln\ x} + c

-------------------------------------------------------------------------------------

However, the integral of x^5 is \frac{1}{6}x^6 + c

This is shown below:

Given that x^5

Applying power rule;

Power rule states that

\int\limits{x^n}\ dx = \frac{x^{n+1}}{n+1} + c

In this case (x^5), n = 5;

So, \int\limits{x^n}\ dx= \frac{x^{n+1}}{n+1} + c

becomes

\int\limits{x^5}\ dx = \frac{x^{5+1}}{5+1} + c

\int\limits{x^5}\ dx = \frac{x^{6}}{6} + c

\int\limits{x^5}\ dx= \frac{x^{6}}{6} + c

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Answer:

  11.6 cm

Step-by-step explanation:

As the page title tells you, the Pythagorean theorem must be applied more than once. As you know, it tells you the square of the hypotenuse is the sum of the squares of the two sides.

  AD² = ED² +EA²

  EA² = AD²-ED² = 7² -6² = 13

  EA = √13 ≈ 3.606

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  CD² = ED² +EC²

  EC² = CD² -ED² = 10² -6² = 64

  EC = √64 = 8

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The length of the horizontal diagonal is ...

  AC = EA +EC = 3.6 +8 = 11.6 . . . cm

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Step-by-step explanation:

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In ΔEFG, the measure of ∠G=90°, GF = 48, EG = 55, and FE = 73. What ratio represents the cosecant of ∠F?
Mrac [35]

Answer:

Cosec <F = 73/55

Step-by-step explanation:

In ΔEFG, the measure of ∠G=90°, GF = 48, EG = 55, and FE = 73. What ratio represents the cosecant of ∠F?

First you must know that;

Cosecant <F = 1/sin<F

Given

∠G=90°, GF = 48, EG = 55, and FE = 73.

ED ,= hyp = 73

EG = opp = 55*side facing <F

Using DOH CAH TOA

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