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patriot [66]
2 years ago
14

Please help me last question

Mathematics
2 answers:
Alona [7]2 years ago
8 0

Answer:

Area of the triangle = 26 cm²

Step-by-step explanation:

The given triangle has the measure of height h = 4 cm

and base of the triangle = 13 cm

We know the formula of the area of a triangle = \frac{1}{2}(Base)(height)

By putting the values in the formula

Area of the triangle = \frac{1}{2}(4)(13)

                                = 2×13

                                = 26 cm²

Therefore, area of the given triangle is 26 cm².

Alborosie2 years ago
7 0

As you can see there are two triangles but it can be done calculating just for one. First we must understand how formula for area of triangle works.

A=\frac{1}{2}bh

Where b represents base (hypotenuse) and h as height of the triangle.

We know that:

b=13cm \\h=4cm

Using this data we fill the formula.

A=\frac{1}{2}\cdot13\cdot4=\frac{13\cdot4}{2}=13\cdot2=\boxed{26cm^2}

Hope this helps.

r3t40

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Answer:  7\text{Ln}\left(e^{2x}+10\right)+C

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=======================================================

Explanation:

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Apply the derivative and multiply both sides by 7 like so

u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\

Integrating leads to

\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of \frac{14e^{2x}}{e^{2x}+10}

4 0
1 year ago
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