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choli [55]
3 years ago
15

Find the hypotenuse (2) of the right triangle.

Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0

Answer:

D) 13m

Step-by-step explanation:

To find the hypotenuse, use the formula for a² + b²= c², where a and b are the values of the triangle's two legs and c is the value of the hypotenuse.

In this case, a² = 7m and b² = 11m. Remember that there is no definite value for a or b, so they can be switched around.

So now our formula will look like this: 7² + 11² = c².

When squaring the two values, you will get a more simplified manner of the equation: 49 + 121 = c².

When we add them altogether, we get 170 = c²

That is not our answer yet, because we want the simplified, and not squared, form as the answer.

In order to do that, we must find the square root of √170, which is 13.0384048104, or just 13 when rounded to the nearest tenth.

So now we know that c², the hypotenuse, is approximately equal to 13.

Just to be sure, plug in all the value into the formula and try again.

  •       7² + 11² = 13².
  •       49 + 121 = 169
  •       170 ≈ 169

Notice: the reason why they are not both is because we rounded 13.0384048104 to 13. But if we square 13.0384048104, we get 170, which is the same value as the one above. Easy, right?

Hope this helps!

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Use linear approximation, i.e. the tangent line, to approximate √16.4 as follows: Let f(x)=√x. Find the equation of the tangent
Ivahew [28]

Answer:

L(x)=\frac{1}{8}x+2\\ L(16.4)=4.05

Step-by-step explanation:

The equation for a tangent line of f(x) in the point (a,f(a)) can be calculated as:

L(x) = f(a) + f'(a)(x-a)

Where L(x) is also call a linear approximation and f'(a) is the value of the derivative of f(x) in (a,f(a)).

So, the derivative of f(x) is:

f(x)=\sqrt{x} \\f'(x)=\frac{1}{2\sqrt{x} }

Then, to find the linear approximation we are going to use the point (16, f(16)). So a is 16 and f(a) and f'(a) are calculated as:

f(16)=\sqrt{16}=4\\f'(16)=\frac{1}{2\sqrt{16} }=\frac{1}{8}

Then, replacing the values, we get that the equation of the tangent line in (16,4) is:

L(x)=4+\frac{1}{8}(x-16)\\L(x) = \frac{1}{8}x+2

Finally, the approximation for \sqrt{16.4} is:

L(16.4)=\frac{1}{8}(16.4)+2\\ L(16.4)=4.05

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