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2 years ago

What's green and fluffy and comes from outer space? I have the answer I just need to know how to graph it

Mathematics

2 answers:

BaLLatris [955]2 years ago

8 0

What’s green and fluffy and comes from outer space?

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1 . 2x – 4y = 8

Rewriting the standard form linear equation into slope intercept form.

-4y = -2x + 8

Dividing both sides by -4.

y = (-2/-4)x + 8/-4

y = (1/2)x - 2

The best match is A and 23.

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2. 2x + 8y = - 64

Rewriting the standard form linear equation into slope intercept form.

8y = - 2x - 64

Dividing both sides by 8.

y = (-2/8)x - 64/8

y = -1/4 x - 8

The best match is N and 19.

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3. -2x + 4y = 8

Rewriting the standard form linear equation into slope intercept form.

4y = 2x + 8

Dividing both sides by 4

y = (2/4)x + 8/4

y = 1/2 x + 2

The best match is M and 4.

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4. 3x + 5y = -15

Rewriting the standard form linear equation into slope intercept form.

5y = - 3x - 15

Dividing both sides by 5.

y = (-3/5)x - 15/5

y = -3/5 x - 3

The best match is M and 15.

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5. 12x - 3y = 3

Rewriting the standard form linear equation into slope intercept form.

-3y = -12x + 3

Dividing both sides by -3.

y = -12/-3 x + 3/-3

y = 4x - 1

The best match is L and 16.

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6. y = - 8

In this case the slope is zero. Which means the rise is zero and run can be anything greater than 1. This will make a horizontal line parallel to the y-axis at the point (0 , -8).

The best match is L and 11.

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7. x + 3y = 18

Rewriting the standard form linear equation into slope intercept form.

3y = -x + 18

Dividing both sides by 3.

y = -1/3 x + 18/3

y = -1/3 x + 6

The best match is T and 2.

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8. -4x + 6y = -54

Rewriting the standard form linear equation into slope intercept form.

6y = - 4x - 54

Dividing both sides by 6.

y = -4/6 x - 54/6

y = -2/3 x - 9

The best match is I and 5.

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9. 4x - y = - 8

Rewriting the standard form linear equation into slope intercept form.

-y = - 4x - 8

Dividing both sides by -1.

y = -4/-1 x - 8/-1

y = 4 x + 8

The best match is B and 22.

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10. 2x - 10y = 20

Rewriting the standard form linear equation into slope intercept form.

-10y = -2x + 20

Dividing both sides by -10.

y = -2/-10 x + 20/-10

y = 1/5 x - 2

The best match is O and 24.

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11. 6x + 9y = 9

Rewriting the standard form linear equation into slope intercept form.

9y = -6x + 9

Dividing both sides by 9.

y = -6/9 x + 9/9

y = -2/3 x + 1

The best match is E and 7.

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12. 5x - 2y = 12

Rewriting the standard form linear equation into slope intercept form.

-2y = -5x + 12

Dividing both sides by -2.

y = -5/-2 x + 12/-2

y = 5/2 x - 6

The best match is A and 8.

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13. 8x + 4y = 40

Rewriting the standard form linear equation into slope intercept form.

4y = -8x + 40

Dividing both sides by 4.

y = -8/4 x + 40/4

y = -2 x + 10

The best match is A and 12.

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14. x + 3y = 27

Rewriting the standard form linear equation into slope intercept form.

3y = -x + 27

Dividing both sides by 3.

y = -1/3 x + 27/3

y = -1/3 x + 9

The best match is W and 2.

____________________________________________________________

ANSWERING TEXT :

A M A R T I A N M E L L O W

23 15 8 22 2 5 12 19 4 7 11 16 24 18

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satela [25.4K]2 years ago

6 0

The <em><u>correct answer</u></em> is:

To graph each of these equations, we will find the x- and y-intercept of each. To find the x-intercept, we replace y with 0 and solve. To find the y-intercept, replace x with 0 and solve.

For 2x-4y=8:

x-intercept:

2x-4(0)=8; 2x-0=8; 2x=8; x=4

y-intercept:

2(0)-4y=8; 0-4y=8; -4y=8; y=-2

Plot these intercepts and draw a line between them.

For 2x+8y=-64:

x-intercept:

2x+8(0)=-64; 2x+0=-64; 2x=-64; x=-32

y-intercept:

2(0)+8y=-64; 0+8y=-64; 8y=-64; y=-8

Plot the intercepts and draw a line between them.

For -2x+4y=8:

x-intercept:

-2x+4(0)=8; -2x+0=8; -2x=8; x=-4

y-intercept:

-2(0)+4y=8; 0+4y=8; 4y=8; y=2

Plot the intercepts and draw a line between them.

For 3x+5y=-15:

x-intercept:

3x+5(0)=-15; 3x+0=-15; 3x=-15; x=-5

y-intercept:

3(0)+5y=-15; 0+5y=-15; 5y=-15; y=-3

Plot the intercepts and draw a line between them.

For 12x-3y=3:

x-intercept:

12x-3(0)=3; 12x-0=3; 12x=3; x=3/12=0.25

y-intercept:

12(0)-3y=3; 0-3y=3; -3y=3; y=-1

Plot the intercepts and draw a line between them.

For y=-8, draw a straight horizontal line at -8.

For x+3y=18:

x-intercept:

x+3(0)=18; x+0=18; x=18

0+3y=18; 3y=18; y=6

Plot the intercepts and draw a line between them.

For -4x+6y=-54:

x-intercept:

-4x+6(0)=-54; -4x+0=-54; -4x=-54; x=13.5

y-intercept:

-4(0)+6y=-54; 0+6y=-54; 6y=-54; y=-9

Plot the intercepts and draw a line between them.

For 4x-y=-8:

x-intercept:

4x-0=-8; 4x=-8; x=-2

y-intercept:

4(0)-y=-8; 0-y=-8; -y=-8; y=8

Plot the intercepts and draw a line between them.

For 2x-10y=20:

x-intercept:

2x-10(0)=20; 2x-0=20; 2x=20; x=10

y-intercept:

2(0)-10y=20; 0-10y=20; -10y=20; y=-2

Plot the intercepts and draw a line between them.

For 6x+9y=9:

x-intercept:

6x+9(0)=9; 6x+0=9; 6x=9; x=1.5

y-intercept:

6(0)+9y=9; 0+9y=9; 9y=9; y=1

Plot the intercepts and draw a line betwen them.

For 5x-2y=12:

x-intercept:

5x-2(0)=12; 5x-0=12; 5x=12; x=2.4

y-intercept:

5(0)-2y=12; 0-2y=12; -2y=12; y=-6

Plot the intercepts and draw a line between them.

For 8x+4y=40:

x-intercept:

8x+4(0)=40; 8x+0=40; 8x=40; x=5

y-intercept:

8(0)+4y=40; 0+4y=40; 4y=40; y=10

Plot the intercepts and draw a line between them.

For x+3y=27:

x-intercept:

x+3(0)=27; x+0=27; x=27

y-intercept:

0+3y=27; 3y=27; y=9

Plot the intercepts and draw a line between them.

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2. 19H+25S=922

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19H+19S=760

Subtract this from 2 to eliminate H,

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Solve for S, then use either equation to solve for H.

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2 years ago

What is the range of the function f(x) = 3x2 + 6x – 8?

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Natali5045456 [20]

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Answer:

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2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago