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Sedaia [141]
3 years ago
13

If the slope of a line is -1 and two points on the line are (-9, 3) and (-10, r), what is the value of r?

Mathematics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

r=4

Step-by-step explanation:

We know that the slope of the line is -1.

We also know that the line passes through (-9, 3) and (-10, r).

We want to find the value of r.

First, let's figure out the equation of our line. We can use the point-slope form:

y-y_1=m(x-x_1)

Where m is the slope and (x₁, y₁) is a point.

So, let's substitute -1 for m. Since we know the point (-9, 3), let's use this for (x₁, y₁).

Substitute:

y-(3)=-1(x-(-9))

Simplify:

y-3=-(x+9)

Distribute:

y-3=-x-9

Add 3 to both sides. So, our equation is:

y=-x-6

It passes through (-10, r) and we want to find the value of r. So, let's substitute -10 for x and r for y, since -10 is our x and r is our y. So:

r=-(-10)-6

Evaluate for r. Distribute:

r=10-6

Subtract:

r=4

So, the value of r is 4.

And we're done!

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Naddik [55]

Answer:

r = -1

Step-by-step explanation:

(y2-y1)/(x2-x1)=m

(r,7)=(x1,y1) and (3,4)=(x2-y2) and m=-3/5

so plug in...

(4-7)/(4-r)=-3/5

then solve for r which will get that r = -1

numerator: 4-7=-3

denominator: 4-r=5 -> 4-5=0+r -> -1=r

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Answer:

law of cosines states c^2=a^2+b^2-2abc

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b^2=676+576-1248cos134°

cos134= -0.6946

b^2=1252-12(-0.6946)

12×(-0.6946)= -8.3352

b^2=1252-(-8.3352)

b^2=1252+8.3352

b^2=1260.33

b=√1260.33

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Kerri has a germen shepherd that has a mass of 30,000 grams. how many kilograms is 30,000 grams
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Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

  • 1st quarter-2
  • 2nd quarter-9
  • 3rd quarter-7
  • 4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4

Hence, there are 4 more home goals than away goals.

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