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user100 [1]
3 years ago
8

What is an equation in point-slope form for the line that contains the points (-1, 4) and (11, -6)?

Mathematics
1 answer:
pshichka [43]3 years ago
5 0

Answer:

The equation in point-slope form is y-4=-\frac{5}{6}(x+1)

Step-by-step explanation:

Given points are;

(-1, 4) and (11, -6)

We will find the slope of the line through the given points.

Slope = m = \frac{y_2-y_1}{x_2-x_1}

m = \frac{-6-4}{11-(-1)}

m = \frac{-10}{12}

m = \frac{-5}{6}

Point slope form of a line is given by;

y-y_1=m(x-x_1)

Putting the point (-1,4) and slope in the equation

y-4=-\frac{5}{6}(x-(-1))\\y-4 =-\frac{5}{6}(x+1)

Hence,

The equation in point-slope form is y-4=-\frac{5}{6}(x+1)

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Which of the following is the solution to 17 - 2x = -11?
Yuliya22 [10]

Answer:

(D)14

Step-by-step explanation:

17-2x=-11

17+11=2x

28=2x

28/2=2x/2

14=x

x=14

8 0
4 years ago
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Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

4 0
3 years ago
Which quadratic inequality does the graph below represent?
sergejj [24]

Answer:

y< or equal to x^2-3.

I hope it helps.

4 0
2 years ago
Need help ASAP please
inysia [295]

Answer:

m=1

Step-by-step explanation:

y2-y1/x2-x1

3 0
3 years ago
What is the answer to this x-7;x=23
blagie [28]

Answer:

16

Step-by-step explanation:

if x=23, 23-7=16

4 0
3 years ago
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