2k/5+3/4=2 please help asap

Part 2: Use the information provided to write the vertex form equation of each parabola.

Serggg [28]

Answer: 1. x = -2(y - 4)² + 1

2. x = -y² + 5

3. y = -5(x + 1)² + 2

<u>Step-by-step explanation:</u>

Notes: The vertex formula of a parabola is x = a(y - k)² + h or y = a(x - h)² + k

(h, k) is the vertex
p is the distance from the vertex to the focus

$\bullet\quad a=\dfrac{1}{4p}$

1)

$\text{Vertex}=(1,4)\qquad \text{Focus}:\bigg(\dfrac{7}{8},4\bigg)\\\\\text{Given}: (h,k)=(1,4)\\\\\\p=focus-vertex=\dfrac{7}{8}-\dfrac{8}{8}=\dfrac{-1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{8})}=\dfrac{1}{-\frac{1}{2}}=-2$

Now input a = -2 and (h, k) = (1, 4) into the equation x = a(y - k)² + h

x = -2(y - 4)² + 1

***********************************************************************************

2)

$\text{Vertex}=(5,0)\qquad \text{Directrix}:x=\dfrac{21}{4}\\\\\text{Given}: (h,k)=(5,0)\\\\\\p=vertex-directrix=\dfrac{20}{4}-\dfrac{21}{4}=\dfrac{-1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{4})}=\dfrac{1}{-1}=-1$

Now input a = -1 and (h, k) = (5, 0) into the equation x = a(y - k)² + h

x = -1(y - 0)² + 5 → x = -y² + 5

***********************************************************************************

3)

$\text{Vertex}=(-1,2)\qquad \text{Directrix}:y=\dfrac{41}{20}\\\\\text{Given}: (h,k)=(-1,2)\\\\\\p=vertex-directrix=\dfrac{40}{20}-\dfrac{41}{20}=\dfrac{-1}{20}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{20})}=\dfrac{1}{-\frac{1}{5}}=-5$

Now input a = -5 and (h, k) = (-1, 2) into the equation y = a(x - h)² + k

y = -5(x + 1)² + 2

8 0

3 years ago

3. What is the slope of the following<br> line?<br> (4,3)<br> 112,-1)]<br> A. - 3<br> B. 4

Likurg_2 [28]

Answer:

2/3

Step-by-step explanation:

that's the answer

.

8 0

3 years ago

Hi, does anyone know how to solve this. If so, please show the working out too. Thanks.

sergiy2304 [10]

See the explanation

<h2>Explanation:</h2><h2 />

I have corrected your diagram so ∅ is the angle at the top of the diagram. In order to solve this problem we have to use Pythagorean theorem and the law of sines. Moreover, I have named two sides as w and z so those variables will help us to solve this problem. So:

The triangle at the bottom is right, so by Pythagorean theorem is true that:

$w^2=4^2+(2\sqrt{2})^2 \\ \\ w^2=24 \\ \\ w=\sqrt{24} \\ \\ w=2\sqrt{6}$

By law of sines:

$\frac{z}{sin\theta}=\frac{w}{sin60^{\circ}} \\ \\ z=\frac{wsin\theta}{sin60^{\circ}} \\ \\ z=\frac{2\sqrt{6}sin\theta}{\sqrt{3}/2} \\ \\ z=4\sqrt{2}sin\theta$

By law of sines again:

$\frac{y}{sin45^{\circ}}=\frac{z}{sin\phi} \\ \\ y=\frac{zsin45^{\circ}}{sin\phi} \\ \\ y=\frac{4\sqrt{2}sin\theta \sqrt{2}/2}{sin\phi} \\ \\ \\ Finally: \\ \\ \boxed{y=\frac{4sin\theta}{sin\phi}}$