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adelina 88 [10]
3 years ago
15

25 POINTS!!!!

Mathematics
2 answers:
erastova [34]3 years ago
5 0
<span>Hey there, Lets solve this problem together. 

You are given two lengths of a right angled triangle, so Pythagoras' theorem is the way to go. 

</span>In algebraic terms, a² + b² = c² where c is the hypotenuse while a and b<span> are the legs of the triangle. 
</span>

In a right angled triangle:
the square of the hypotenuse is equal to
the sum of the squares of the other two sides.

c² (the diagonal squared) = a² (one length squared) + b²<span> (the other length squared)  

</span><span>leg a and b are both 3 

</span>3^2 + 3^2 = c^2&#10;

= 3 * 3 = 3^2&#10;

= 9 + 9 = c^2 

9 + 9 =18 

18 = c^2 

<span>The opposite of ^2 is ^1/2 also known as the square root
</span>
<span>c = sqrt(18)</span>

Solved = <span>c≈4.24 
</span>



postnew [5]3 years ago
3 0
It will be square root of 6
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A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seco
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answer:

f(t) = -9.9788169(t -0.76)^2 +57

Step-by-step explanation:

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the  number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

now we know if a rock is thrown it moves in a shape of a parabola which we see this equation is quadratic. Now we will use the turning point equation for a quadratic equation to get a equation for the height which the format is f(x)= a(x-p)^2 +q  , where (p,q) is the turning point. now we substitute the turning point

f(t) = a(t-0.76)^2 + 57, now we will substitute the other point on the graph or on the function that we found which is (3.15, 0) then solve for a.

0 = a(3.15 - 0.76)^2 + 57

-57 =a(2.39)^2  

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-57/5.7121 =a

-9.9788169 = a      then we substitute a to get the quadratic equation therefore f is

f(t) = -9.9788169(t-0.76)^2 +57

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