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4 years ago

Write an equation in slope intercept form for the line that travels through (1,-2) and (6,8)

Mathematics

1 answer:

MariettaO [177]4 years ago

3 0

Answer:

y=2/1x

Step-by-step explanation:

Subtract (6,8) by (1, -2)

8--2/6-1=10/5

Simply

10/5=2

Plug in

y=2x

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I needs help with it

Slav-nsk [51]

Answer:

a) initial height = 0 m

b) 8 s

c) vertex = (4, 32)

maximum height = 32 m

Step-by-step explanation:

Given equation:

$h=-2t^2+16t$

where:

h is the height of the arrow about the ground (in metres)
t is the time (in seconds)

The initial height of the arrow will be at the beginning of its journey, so when t = 0

Substitute t = 0 into the equation and solve for h:

$\implies h=-2(0)^2+16(0)$

$\implies h=0$

Therefore, the initial height of the arrow is 0 m.

The arrow will hit the ground when the height is 0 m.

Substitute h = 0 into the given equation:

$\implies -2t^2+16t=0$

Factor out -2t:

$\implies -2t(t-8)=0$

Therefore:

$\implies -2t=0 \implies t=0$

And:

$\implies t-8=0 \implies t=8$

Therefore, the arrow will hit the ground at 8 seconds.

The vertex is the turning point of a parabola - it's minimum or maximum point.

As the leading coefficient for the given equation is negative, the parabola opens downwards and so its vertex is its maximum point.

There are different ways to find the vertex of a parabola. The easiest way to find the x-coordinate is by calculating the midpoint of the x-intercepts.

We have determined that the arrow has a height of 0 m at 0 seconds and 8 seconds. Therefore, t = 0 and t = 8 are the x-intercepts. The midpoint is halfway between zero and 8, so the midpoint is t = 4

To find the y-coordinate, substitute t = 4 into the equation and solve for h:

$\implies h=-2(4)^2+16(4)$

$\implies h=-32+64$

$\implies h=32$

Therefore, the coordinates of the vertex are (4, 32)

The maximum height is the y-coordinate of the vertex.

Therefore, the maximum height is 32 m

6 0

2 years ago

Mario’s younger brother is 2 3/4 tall. Mario is 1 3/4 feet taller than his younger brother. Which number line represents Mario’s

harkovskaia [24]

Answer:

Graph A - 4 1/2 feet

Step-by-step explanation:

Since Mario is taller that his younger brother - add the two heights together.

2 3/4 + 1 3/4 =

I added the whole numbers first 2 + 1 = 3; then the fractions 3/4 + 3/4 = 1 1/2

Add 3 + 1 1/2 = 4 1/2

6 0

4 years ago

Help me ASAP Helppppppppppppppppp

Natalija [7]

Answer:

(2,4)

Step-by-step explanation:

I painfully counted every square to find that a 1 by 1 square is divided into an 8 by 8 square.

K is one square above 1/2, for the second coordinate is 4.

K is also 2 to the left of 0, so the first coordinate is 2.

Therefore, the coordinates of K are (2,4).

6 0

3 years ago

Find the missing side or angle for each variable round your answer to the nearest tenth

Ray Of Light [21]

Answer:

x is about 37.6 degrees

Step-by-step explanation:

hope that helped

6 0

3 years ago

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

Given α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78

Part 2A:

Given α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68


Part 3A:

Given α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107


Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097

5 0

3 years ago