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4 years ago

9

I need some help on this with my geometry stuff

1 answer:

AleksandrR [38]4 years ago

7 0

Length of AA' = √(5^2 + 2^2) = √29 = 5.39

Answer

5.39

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Suppose that you see a flock of birds at an angle of elevation of 32 degrees if the birds are at an altitude of 12,000 feet and

Firlakuza [10]

We know that

tan 32°=opposite side/adjacent side
[adjacent side]=[opposite side]/tan 32°

in this problem

opposite side-----> will be the altitude of the flock of birds minus your eye level above the ground
(12000-6)-----> 11994 ft

adjacent side------->will be the horizontal distance between you and the birds
32°------> <span>an angle of elevation
</span>
[adjacent side]=[11994]/tan 32°-------> 19194.41 ft

the answer is
<span>The horizontal distance from the birds is 19194.41 ft</span>

7 0

3 years ago

Binomial numbers, Stifel

Ivahew [28]

Answer:

binomiales, números combinatorios o combinaciones son números estudiados en combinatoria que corresponden al número

Step-by-step explanation:

5 0

3 years ago

Solve for x.<br> 6(х – 1) = 9(х + 2)<br> Ox=-8<br> 0 x = -3<br> Ox= 3<br> Ox= 8

Maru [420]

The answer to you’re question would be A. x = -8

3 0

3 years ago

Find lim h->0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925

Svetach [21]

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h$

Carry out the binomial expansion in the numerator:

$(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4$

Then the 9⁴ terms cancel each other, so in the limit we have

$\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h$

Since <em>h</em> is approaching 0, that means <em>h</em> ≠ 0, so we can cancel the common factor of <em>h</em> in both numerator and denominator:

$\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)$

Then when <em>h</em> converges to 0, each remaining term containing <em>h</em> goes to 0, leaving you with

$\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}$

or choice C.

Alternatively, you can recognize the given limit as the derivative of <em>f(x)</em> at <em>x</em> = 9:

$f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h$

We have <em>f(x)</em> = <em>x</em> ⁴, so <em>f '(x)</em> = 4<em>x</em> ³, and evaluating this at <em>x</em> = 9 gives the same result, 2916.

8 0

3 years ago

How to solve 2^3 * 2^2?

const2013 [10]

(2x2x2)(2x2)
=2x2x2x2x2
=2x2x2x2x2
=32

5 0

4 years ago

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