MY Chaim tried to prove that Triangle JKL = Triangle LMN.

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{

il63 [147K]

Answer:

$\frac{x-3}{2x-3}$ . hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

$x^2-5x+6$

Now let's find what ac is (it's just c since a=1...)

$AC= 6$

List factors of -6

$\pm1, \pm2, \pm3, \pm6$ .

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

$(x-2)(x-3)$

Ok now let's factor the second equation:

$2x^2-7x+6$

Multiply a and c

$AC = 12$

List factors of 12:

$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ .

Factors that add up to -7 and multiply to 12:

$-3\ and\ -4$

Rewrite equation:

$2x^2-4x-3x+6$

Group terms:

$(2x^2-4x)+(-3x+6)$

Factor out GCF:

$2x(x-2)-3(x-2)$

Rewrite:

$(2x-3)(x-2)$

Now let's write out the equation using these factors:

$\frac{(x-2)(x-3)}{(2x-3)(x-2)}$ .

Here we can factor out the x-2 and the simplified form is:

$\frac{x-3}{2x-3}$

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0

2 years ago

Medal will be awarded:. When constructing an inscribed polygon with a compass and straightedge, how should you start the constru

Ymorist [56]

The draw an inscribed polygon inside the circle, the very thing to do is to draw the circle. Thus, the answer is the fourth choice, "Place a point on your paper and then use a compass to construct a circle". After which, we can locate the vertices of the polygon in the circumference of the circle and connect them to make the polygon.

6 0

3 years ago

Points (1,3), (-4,7), and (-29,K) are collinear. Find K. How do i solve this?

oee [108]

Co linear points have same slope
Points (1,3), (-4,7), and (-29,K)
m1,2=7-3/-4-1=-4/5

m2,3=K-7/-29+4
m2,3=K-7/-25

-4/5=K-7/-25
4/5=K-7/25
4=K-7/5
20=K-7
K=27

3 0

3 years ago

A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How

sergeinik [125]

24ft/s for the 1st second then it'd be 9.8mp/s which is the speed of velocity towards earth, so the 24ft/s - the 9.8 would be 14.2ft/s which is the second, the third would then be 4.4 after again subtracting 9.8. That is as far as I got, hope that helps even slightly

7 0

3 years ago

Question 1: The water level in a tank can be modeled by the function h(t)=4cos(+)+10, where t is the number of hours

Amanda [17]

Answer:

Question 1: The hours that will pass between two consecutive times, when the water is at its maximum height is π hours

Question 2: Sin of the angle is -0.8

Step-by-step explanation:

Question 1: Here we have h(t) = 4·cos(t) + 10

The maximum water level can be found by differentiating h(t) and equating the result to zero as follows;

$\frac{\mathrm{d} h(t)}{\mathrm{d} t} = \frac{\mathrm{d} \left (4cos(t) + 10 \right )}{\mathrm{d} t} = 0$

$\frac{\mathrm{d} h(t)}{\mathrm{d} t} = - 4 \times sin(t) = 0$

∴ sin(t) = 0

t = 0, π, 2π

Therefore, the hours that will pass between two consecutive times, when the water is at its maximum height = π hours.

Question 2:

B = (3, -4)

Equation of circle = x² + y² = 25

Here we have

Distance moved along x coordinate = 3

Distance moved along y coordinate = -4

Therefore, we have;

$Tan \theta = \frac{Distance \ moved \ along \ y \ coordinate}{Distance \ moved \ along \ x \ coordinate} = \frac{-4}{3}$

$\therefore \theta = Tan^{-1}(\frac{-4}{3}) = -53.13^{\circ}$

Sinθ = sin(-53.13) = -0.799≈ -0.8.

5 0

3 years ago