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2 years ago

9

What's the work and answer for this problem with the Pythagorean theorem

2 answers:

Novay_Z [31]2 years ago

7 0

Answer:

b = 3

Step-by-step explanation:

$a^2+b^2=c^2$

$12^2+b=15^2$

$15^2 - 12^2 = 3$

$b = 3^2$

hope this helps :)

Rus_ich [418]2 years ago

7 0

Answer:

the answer is b=3

hope this helped bye

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The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be w

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Answer:

61

Step-by-step explanation:

Let's find the points $A$ and $B$ .

We know that the $y$ -coordinates of both are $3$ .

So let's first solve:

$3=4x^2+x-1$

Subtract 3 on both sides:

$0=4x^2+x-1-3$

Simplify:

$0=4x^2+x-4$

I'm going to use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , to solve.

We must first compare to the quadratic equation, $ax^2+bx+c=0$ .

$a=4$

$b=1$

$c=-4$

$\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}$

$\frac{-1 \pm \sqrt{1+64}}{8}$

$\frac{-1 \pm \sqrt{65}}{8}$

Since the distance between the points $A$ and $B$ is horizontal. We know this because they share the same $y-coordinate$ .This means we just need to find the positive difference between the $x$ -values we found for the points of $A$ and $B$ .

So that is, the distance between $A$ and $B$ is:

$\frac{-1+\sqrt{65}}{8}-\frac{-1-\sqrt{65}}{8}$

$\frac{-1+\sqrt{65}+1+\sqrt{65}}{8}$

$\frac{2\sqrt{65}}{8}$

$\frac{\sqrt{65}}{4}$

If we compare this to $\frac{\sqrt{m}}{n}$ , we should see that:

$m=65 \text{ and } n=4$ .

So $m-n=65-4=61$ .

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