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seropon [69]
3 years ago
8

1. Calcula quatre solucions diferents de l'equació x – y = 5 , representa els punts amb ajuda d'uns eixos, i comprova que tots e

stan sobre una mateixa recta

Mathematics
1 answer:
Bess [88]3 years ago
3 0

Answer:

Hence 4 different points are  as

A(1,-4) ,B(2,-3)  ,C(3,-2)  D(4,-1)

Step-by-step explanation:

Given:

A line  represents as x-y=5

To Find:

4 different points on that line with.

Solution:

Now calculate the points on both axes in order to draw the line .

So Put x=0 then y=-5

And when y= 0, x=5

Two points are (0,-5) and (5,0)

Hence draw line through this point  and obtain the other two points by plotting on graph as follows:

(Refer the attachment)

Here the line is  4th quadrant so x will positive and y will negative

And addition of both should result in 5

1) Consider As x=1 then,

y=x-5=1-5=-4

y=-4

Point will be (1,-4)

2)consider As x=2 then

y=x-5=2-5=-3

y=-3

Point will be (2,-3)

3)Consider As x=3 then

y=x-5=3-5= -2

y=-2

Point will be (3,-2)

4) Consider AS x=4 then

y=x-5=4-5=-1

y=-1

Point will be (4,-1)

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3 years ago
A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40
malfutka [58]
PART A

The given equation is

h(t) = - 16 {t}^{2} + 40t + 4

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4

We add and subtract

- 16(- \frac{5}{4} )^{2}

to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4

We again factor -16 out of the first two terms to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4

This implies that,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4

The quadratic trinomial above is a perfect square.

h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4

This finally simplifies to,

h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29

The vertex of this function is

V( \frac{5}{4} ,29)

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

29

PART B

When the ball hits the ground,

h(t) = 0

This implies that,

- 16 ( t- \frac{5}{4}) ^{2} +29 = 0

We add -29 to both sides to get,

- 16 ( t- \frac{5}{4}) ^{2} = - 29

This implies that,

( t- \frac{5}{4}) ^{2} = \frac{29}{16}

t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }

t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}

t = \frac{ 5 + \sqrt{29} }{4} = 2.60

or

t = \frac{ 5 - \sqrt{29} }{4} = - 0.10

Since time cannot be negative, we discard the negative value and pick,

t = 2.60s
8 0
3 years ago
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