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taurus [48]
3 years ago
11

Pleaseeeeee solveee 25 points pleaze help geometry

Mathematics
1 answer:
tatiyna3 years ago
8 0

Answer:

(-3/2, 6)

Step-by-step explanation:

(-3,8) (6,-4)

As we move along the line segment with endpoints (-3,8) and (6,-4) from  (-3,8) to (6,-4) x increases by 9 units (from -3 to 6) and y decreases by -12 units (from 8 to -4).

for ratio of 1:5

Since 1+5=6, the point we are looking for is 1/6 of the way from (-3,8) to (6,-4).  

So, the desired point is

(-3 + (1/6)(9), 8 - (1/6)(12) )

= (-3 + 3/2 , 8 - 2)

= ( (-6 + 3)/2, 6)

= (-3/2, 6)

for map reference see the image below

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Subtract. Write your answer in simplest form.
vlabodo [156]
Answer: C

Work:

1) Change the Mixed number into an improper fraction
3 x 8 = 24
1+24 = 25/3

2) Get both fractions on equal terms. Find the Greatest Common Denominator. In this case, it’s 24. Multiply by what you need to get 24. 8x3=24 and 12x2=24

3) Multiply Accordingly
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4) Subtract and change back to Mixed Number
75-14 = 61
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3 0
2 years ago
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Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).
guajiro [1.7K]

The area of the triangle with the vertices (0, 5), (2, -2), and (5, 1) by using the calculus is  21 square unit.

We need to find the equation among all possible pairs and then integrate the equations from one co-ordinate to another co-ordinate

Equation of line passing through (0,5) and (2,-2) is

y-5 = [(-2-5)/(2-0)](x-0)

=>y-5 = (-7) /2x

=>y= (-7/2x)+5 -------(eq1)

Equation of line passing through (0,5) and (5,1) is

y-5 = [(1-5) / (5-0)](x-0)

=>y-5 = (-4/5)x

=>y = (-4/5)x+5-------(eq2)

Equation of line passing from (2,-2) and (5,1) is

y-(-2) = [[1 - (-2)] / (5-2)] / (x-2)

=>y+2=(3/3)(x-2)

=>y=x-4--------(eq3)

Now, we use definite integration to find the area between the different equation of line.

So, area enclosed between the equations is given by the

area =\int\limits^5_2[(-4/5)x+5 - (-7/2)x + 5)dx  + \int\limits^5_1[(-4/5)x+5 -(x-4)]dx

=>area=\int\limits^5_2(7/2-4/5)x dx + \int\limits^5_1((-9/5)x+9)dx

Using properties of integration,\int\limits x\, dx=x^{2}/2

=>area=\int\limits^5_2(27/10)x dx + \int\limits^5_1(-9/5)x+9)dx

=>area=([27/10)×[5² - 2²])/2 + [ (-9/5)×(5²-1²) ]/2 +9×(5-1)

=>area=(27/20)×(25-4) + (-9/5)×24+9×4

=>area = (27×21)/20 + (-216)/5+ 36

=>area=(567/20) - (216/5) + 36

=>area= [(567-261×4)+(36×20)]/20

=>area=[(567-864)+720]/20

=>area=423/20

=>area=21 square unit.

Hence, area of triangle is 21 square unit.

To know more about area of triangle, visit here:

brainly.com/question/19305981

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4 0
1 year ago
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