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Temka [501]
3 years ago
11

Diagram below shows the curve of a quadratic function

Mathematics
1 answer:
uranmaximum [27]3 years ago
3 0

Answer:

q = -8, k = 2.

r = -6.

Step-by-step explanation:

f(x) = (x - p)^2 + q

This is the vertex form of a quadratic where the vertex is at the point (p, q).

Now the x intercepts are at -6 and 2 and the curve is symmetrical about the line x = p.

The value of p is the midpoint of  -6 and 2 which is (-6+2) / 2 = -2.

So we have:

f(x) = 1/2(x - -2)^2 + q

f(x) = 1/2(x + 2)^2 + q

Now the graph passes through the point (2, 0) , where it intersects the x axis, therefore, substituting x = 2 and f(x) = 0:

0 = 1/2(2 + 2)^2 + q

0 = 1/2*16 + q

0 = 8 + q

q = -8.

Now convert this to standard form to find k:

f(x) = 1/2(x + 2)^2 - 8

f(x) = 1/2(x^2 + 4x + 4) - 8

f(x) = 1/2x^2 + 2x + 2 - 8

f(x) = 1/2x^2 + 2x - 6

So k = 2.

The  r  is the y coordinate when x = 0.

so r = 1/2(0+2)^2 - 8

=  -6.

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what is 3.73/33.46? the question is jose bought a magazine for 3.73 and some light bulbs for 4.78. he spent a total of 37.19. ho
bekas [8.4K]

Answer:

He bought 7 lightbulbs.

Step-by-step explanation:

37.19 - 3.73 = 33.46

33.46/4.78=7

3 0
3 years ago
Please help me!!!!!!!!!!!!!!!!!!!!!!!!
romanna [79]

Answer:

a. 81.9

b. 144.7

c. 110.2

d. 48

Step-by-step explanation:

a. 120/109.2 = 90/x

x = 81.9

b. 45.6/120 = 55/x

x = 144.7

c. 304.8/96 = 350/x

x = 110.2

d. 109.2/45.6 = 115/x

x = 48

5 0
3 years ago
L1 : y = 2x , (2) find the equation of the line L2 perpendicular to L1 passing through the point P = (1, 2).
just olya [345]

Answer:

<h2>2y+x = 5</h2>

Step-by-step explanation:

Given the line L1 as y = 2x perpendicular to an unknown line L2 passing through the point P = (1, 2), we are to find the equation of line L2. to find the equation of the line L2, we will use the point-slope equation of a line expressed as y-y₀ = m(x-x₀)

m is the slope of the unknown line

(x₀, y₀) is the given point.

First is to get the slope of the known line:

comparing the line L1: y = 2x with the standard equation of the line y = mx+c, it can be seen that m = 2

Then we will calculate the slope of the required line.

Since L1 is perpendicular to L2, the product of their slope will be -1 i.e

mm₁ = -1 where m₁ is the slope of the required line L2.

Given m =2

m₁ = -1/m

m₁ = -1/2

Finally we will calculate the equation of line L2 by substituting the slope of line L2 and the point in the point slope equation above;

y-y₀ = m(x-x₀)

Given (x₀, y₀) = (1,2) and m₁ = -1/2

y-2 = -1/2(x-1)

open the parenthesis

y-2 = -x/2+1/2

multiply through by 2:

2y-4 = -x+1

2y+x = 1+4

2y+x = 5

<em>Hence the equation of the line L2 is 2y+x = 5</em>

<em></em>

8 0
2 years ago
The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, w
Aloiza [94]

Answer:

<h2>z=36</h2>

Step-by-step explanation:

According to the question,

x∝y^3          .......(1)

y∝\sqrt{z}    .......(2)

From equation 1,2 let constant of proportionality be k1,k2 respectively.

⇒x=k1(y^3)            .......(3)

⇒y=k2(\sqrt{z} )    .......(4)

From the above equations putting 4 into 3,

x=k1((k2\sqrt{z})^3) =k1.k2^3.(\sqrt{z})^3

Let the new constant to the above equation be k3,

x=k3(\sqrt{z})^3

Given,if x=1, when z=4

1=k3(\sqrt{4} )^3=k3(8)

⇒k3=\frac{1}{8}

Now if x=27, then z=?

27=\frac{1}{8} (\sqrt{z} )^3

⇒(\sqrt{z} )^3=27(8)

⇒\sqrt{z}=3(2)=6

z=36

4 0
3 years ago
Factor completely.<br>........
posledela
(2x+5)(5x+1) is completley factored
4 0
3 years ago
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