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kobusy [5.1K]
3 years ago
12

fine the y intercepts and intersection point for each graph. then write a system of equation for each graph.​

Mathematics
1 answer:
vovikov84 [41]3 years ago
6 0

Answer:

In the previous lesson, you learned how to graph points on the coordinate plane. We can connect two points with a straight line.

To graph the equation of a line, we plot at least two points whose coordinates satisfy the equation, and then connect the points with a line. We call these equations "linear" because the graph of these equations is a straight line.

There are two important things that can help you graph an equation, slope and y-intercept.

Slope

We're familiar with the word "slope" as it relates to mountains. Skiers and snowboarders refer to "hitting the slopes." On the coordinate plane, the steepness, or slant, of a line is called the slope. Slope is the ratio of the change in the y-value over the change in the x-value. Carpenters and builders call this ratio the "rise over the run." Using any two points on a line, you can calculate its slope using this formula.

Let's use these two points to calculate the slope m of this line.

A = (1,1) and B = (2,3)

Subtract the y value of point A from the y-value of point B to find the change in the y value, which is 2. Then subtract the x value of point A from the x value of point B to find the change in x, which is 1. The slope is 2 divided by 1, or 2.

When a line has positive slope, like this one, it rises from left to right.

WATCH OUT! Always use the same order in the numerator and denominator!

It doesn't really matter whether you subtract the values of point A from the values of point B, or the values of point B from the values of point A. Try it - you'll get the same answer both ways. But you must use the same order for both the numerator and denominator!

You can't subtract the y value of point A from the y value of point B, and the x value of point B from the x value of point A - your answer will be wrong.

Let's look at another line. This line has a negative slope, it falls from left to right. We can take any two points on this line and find the slope. Let's take C (0, -1) and D (2, -5).

Using these two points, we can calculate the slope of this line. We subtract the y value of point C from the y value of point D, and the x value of point C from the x value of point D, and divide the first value by the second value. The slope is -2.

Y-Intercept

There's another important value associated with graphing a line on the coordinate plane. It's called the "y intercept" and it's the y value of the point where the line intersects the y- axis. For this line, the y-intercept is "negative 1." You can find the y-intercept by looking at the graph and seeing which point crosses the y axis. This point will always have an x coordinate of zero. This is another way to find the y-intercept, if you know the equation, the y-intercept is the solution to the equation when x = 0.

Equations

Knowing how to find the slope and the y-intercept helps us to graph a line when we know its equation, and also helps us to find the equation of a line when we have its graph. The equation of a line can always be written in this form, where m is the slope and b is the y-intercept:

y = mx + b

Let's find the equation for this line. Pick any two points, in this diagram, A = (1, 1) and B = (2, 3).

We found that the slope m for this line is 2. By looking at the graph, we can see that it intersects the y-axis at the point (0, –1), so –1 is the value of b, the y-intercept. Substituting these values into the equation formula, we get:

y = 2x –1

The line shows the solution to the equation: that is, it shows all the values that satisfy the equation. If we substitute the x and y values of a point on the line into the equation, you will get a true statement. We'll try it with the point (2, 3).

Let's substitute x = 2 and y = 3 into the equation. We get "3 = 3", a true statement, so this point satisfies the equation of the line.

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Answer:

The linear equation is;

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Step-by-step explanation:

Whereby we have the following relation;

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Answer:

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Step-by-step explanation:

Let

Claire stickers= x

Josh = 2x

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Total stickers = Claire stickers + Josh stickers + Ann stickers

157 = x + (2x) + (2x+7)

157 = x + 2x + 2x + 7

157 = 5x + 7

Subtract 7 from both sides

157 - 7 = 5x + 7 - 7

150 = 5x

Divide both sides by 5

150/5 = x

30 = x

x = 30

Claire stickers= x

=30 stickers

Josh = 2x

= 2(30)

= 60 stickers

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= 2(30) + 7

= 60 + 7

= 67 stickers

Claire = 30 stickers

Josh = 60 stickers

Ann = 67 stickers

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A dairy company gets milk from two dairies and then blends the milk to get the desired amount of butterfat. Milk from dairy I co
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Answer:

a) i The company should buy 40 gallons from dairy I and 60 gallons from dairy

ii) What is the maximum amount of​ butterfat? The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b.The excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

Step-by-step explanation:

a. How much milk from each supplier should the company buy to get at most 100 gallons of milk with the maximum amount of​ butterfat?

From the question, we are told that:

Milk from dairy I costs ​$2.40 per​ gallon, Milk from dairy II costs ​$0.80 per gallon.

Let's represent:

Number of gallons of Milk from dairy I = x

Number of gallons of Milk from dairy II = y

At most ​$144 is available for purchasing milk.

$2.40 × x + $0.80 × y = 144

2.40x + 0.80y = 144........ Equation 1

x + y = 100....... Equation 2

x = 100 - y

2.40(100 - y) + 0.80y = 144

240 - 2.4y + 0.80y = 144

-1.60y = 144 - 240

-1.6y = -96

y = -96/-1.6

y = 60

From Equation 2

x + y = 100....... Equation 2

x + 60 = 100

x = 100 - 60

x = 40

Therefore, since number of gallons of Milk from dairy I = x and number of gallons of Milk from dairy II = y

The company should buy 40 gallons from dairy I and 60 gallons from dairy

II. What is the maximum amount of​ butterfat?

From the question

Dairy I can supply at most 50 gallons averaging 3.9​% ​butterfat,

50 gallons = 3.9% butterfat

40 gallons =

Cross Multiply

= 40 × 3.9/50

= 3.12%

Dairy II can supply at most 90 gallons averaging 2.9​% butterfat.

90 gallons of milk = 2.9% butter fat

60 gallons of milk =

Cross Multiply

= 60 × 2.9%/90

=1.9333333333%

≈ 1.93%

The total amount of butterfat from Diary I and Diary II = 3.12% + 1.93%

=5.05%

b. The solution from part a leaves both dairy I and dairy II with excess capacity. Calculate the amount of additional milk each dairy could produce.

Excess capacity of Diary I =

50 gallons - 40 gallons = 10 gallons

Excess capacity of Diary II =

90 gallons - 60 gallons = 30 gallons

Therefore, the excess capacity of dairy I is 10 ​gallons, and for dairy II it is 30 gallons.

3 0
3 years ago
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