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anyanavicka [17]
3 years ago
6

Find the slope of the line that goes through the points (-3, 6) and (-4,-1)

Mathematics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

7

Step-by-step explanation:

1) (-3,6) and (-4, -1) go into this formula m = (y2 - y1) / (x2 - x1)

2) -1 - 6 / -4 - (-3) = -7/-1

3) -7/-1 = 7

4) m=7

m = slope

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answer: the missing side is 5

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A team of biologists captured and tagged 65 deer in a forest. Two weeks later, the
Gre4nikov [31]

Answer:

1/6x= 65

Step-by-step explanation:

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3 years ago
Write the equation of the line in point-slope form that passes through (5, -1) and has a slope of 2/3
kolezko [41]

Answer:

y-(-1)=2/3(x-5)

Step-by-step explanation:

Point slope form is

y-y_1=m(x-x_1) where m is the slope, x_1 is the x coordinate of a point that the line passes through, and y_1 is the y coordinate of a point the line passes through.

Substituting, we get

y-(-1)=2/3(x-5)

5 0
3 years ago
The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.
Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m  + 40 m  + c = 120 m

61 m +  c = 120 m

c = 120 m - 61 m

c = 59 m

                         B

                     ↓            ↓  

                  ↓                          ↓

                ↓                                       ↓

            A →  →  → →  →  → →  → →    →    →  C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0

Angle B is calculated as follows;

Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0

Angle C is calculated as follows;

Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0

Therefore, the largest angle of the field is 149⁰.

       

8 0
3 years ago
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