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3 years ago

Quadratic formula for x^2-4x+5=0

1 answer:

3 0

Here go to this website, it will explain everything
https://www.symbolab.com/solver/quadratic-equation-calculator/2x%5E%7B2%7D-4x+5=0

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The diagram shows the distance

worty [1.4K]

Rjdnkdkdkdkkddjjfnfnfnfnfmfmfkkffkfkkffkmfmfjfjfkt

6 0

2 years ago

Emma received a $40 gift card that could be used to download television programs. After downloading 5 programs, she had $30 rema

aalyn [17]

Answer:

x= 1 program

x= $2

Step-by-step explanation:

40-30= 10 <- what she spent

10/5 = 2

5 = Amount of Programs

Each program is $2.

Your welcome :)

8 0

3 years ago

A parabola has a focus of F(2, -0.5) and a directrix of y=-1.5 P(x,y) represents any point on the parabola, while D(x, -1.5) rep

prohojiy [21]

The sketch of the parabola is attached below

We have the focus $(a,b) = (2, -0.5)$
The point $P(x,y)$
The directrix, c at $y=-1.5$

The steps to find the equation of the parabola are as follows

Step 1
Find the distance between the focus and the point P using Pythagoras. We have two coordinates; $(2, -0.5)$ and $(x,y)$ .
We need the vertical and horizontal distances to find the hypotenuse (the diagram is shown in the second diagram).
The distance between the focus and point P is given by
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$

Step 2
Find the distance between the point P to the directrix $c$ . It is a vertical distance between y and c, expressed as $y-c$

Step 3
The equation of parabola is then given as
$\sqrt{ (x-a)^{2}+ (y-b)^{2} }$ = $y-c$
$(x-a)^{2}+ (y-b)^{2}= (y-c)^{2}$ ⇒ substituting a, b and c
$(x-2)^{2}+ (y--0.5)^{2} = (y--1.5)^{2}$
$(x-2)^{2}+ (y+0.5)^{2}= (y+1.5)^{2}$ ⇒Rearranging and making $y$ the subject gives

$y= \frac{ x^{2} }{2} -2x+1$

7 0

3 years ago

Please help me with this homework

snow_lady [41]

Answer:

10.44 m

Step-by-step explanation:

Solve this as a triange. The base is 3m and the perpendicular is 10m. The angle between the base and wall is 90° so it will be a right angled triangle.

Solve this by Pythagoras theorm:

let length of ladder is c

base is b

And perpendicular is a

Acc to Pythagoras theorem:

C² = a² + b²

C² = 10² + 3²

C² = 100 + 9

C² = 109

√C² = √109

C = 10.44m

8 0

2 years ago

Find the equation of a line parallel to −5x+y=2 that contains the point (5,−2). Write the equation in slope-intercept form.

mario62 [17]

Answer:

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

Step-by-step explanation:

Parallel lines have same slopes.

Line 1, L1: 5x-2y=20 is in standard form Ax+By=C therefore slope m1= -A/B = -5/-2 = 5/2 or you can solve it for y so you will have the equation in slope-intercept form.

5x-2y = 20

-2y = -5x+20

y = (-5/-2)x+20/(-2)

y = (5/2)x-10 hence m1=5/2 and y-intercept is -10

Line 2 , L2: y-y1 = m (x-x1), m=m2=m1=5/2

Point p(5,0) or p(x1,y1) therefore x1=5 , y1=0 and m=5/2

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

5 0

2 years ago