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tekilochka [14]
3 years ago
6

PY=2, HP=3, find HY Geometry

Mathematics
1 answer:
klemol [59]3 years ago
5 0

Answer:

HY=\sqrt{5}\ units

Step-by-step explanation:

we know that

The triangle HPY is a right triangle (because the angle ∠HYP is a right angle)

Applying the Pythagorean Theorem

HP^2=HY^2+PY^2

substitute the given values

3^2=HY^2+2^2

Solve for HY

9=HY^2+4

HY^2=9-4

HY^2=5

HY=\sqrt{5}\ units

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Explanation:

1) 2x^2 -5x -3 = 0 . . . . standard form equation

To convert this to factored form, you can look for factors of the product (2)(-3) that have a sum of -5. It can help to start by listing the ways that -6 can be factored. Since we want the sum of factors to be negative, we want to have larger negative factors.

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The sums of these factor pairs are -5 (what we want) and -1 (not relevant). We can call these factors p=1 and q=-6.

If a = 2 is the leading coefficient of our standard form quadratic, we want to use these factors in the form ...

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  (2x +1)(2x +(-6))/2 . . . .fill in the values we know

  (2x +1)(x -3) . . . . . . .  factor 2 from the second binomial

So, the factored form of the quadratic equation is ...

  (2x +1)(x -3) = 0 . . . . factored form equation

__

2) f(x) = x^2 +7x +10 . . . . standard form quadratic function

Using the thinking process described above, we are looking for factors of 10 that have a sum of 7. We know those are 2 and 5. So, the factored form of the function is ...

  f(x) = (x +2)(x +5) . . . . . . factored form quadratic function

The leading coefficient is 1, so we have no further work to do.

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The graph attached below shows this function crosses the x-axis when x=-2 and x = -5. These values of x are variously called "roots", "x-intercepts", and "zeros" of the function. They are values for which the factors and the function are zero. (x+2=0 when x=-2, for example)

<u>Solutions</u>

Often, we are interested in solving the equation ...

  f(x) = 0

For that equation, the <em>solutions</em> <u>are</u> the <em>zeros</em> or <em>x-intercepts</em> or <em>roots</em>. The graph attached also shows solutions for ...

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