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3 years ago

Fellow 9. What is the distance between (-3, 4) and (5, 1)?

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

7 0

Answer: The exact distance is $\sqrt{73}$ units

The approximate distance is roughly 8.5440037 units

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Explanation:

Use the distance formula

$(x_1,y_1) = (-3,4)\\\\(x_2,y_2) = (5,1)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-3-5)^2 + (4-1)^2}\\\\d = \sqrt{(-8)^2 + (3)^2}\\\\d = \sqrt{64 + 9}\\\\d = \sqrt{73}\\\\d \approx 8.5440037\\\\$

As an alternative, you can use the pythagorean theorem.

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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

Consider the equation 3a+0.1n=123. If we solve this equation for n, which equation would result? A. 0.1n=123-3a B. N=123-3a-0.1

neonofarm [45]

Answer:

n = (123 - 3a) / 0.1

Step-by-step explanation:

Given:

3a + 0.1n = 123

Solve for n

3a + 0.1n = 123

Subtract 3a from both sides

3a + 0.1n - 3a = 123 - 3a

0.1n = 123 - 3a

Divide both sides by 0.1

n = (123 - 3a) / 0.1

The resulting equation if 3a + 0.1n = 123 is solved for n is n = (123 - 3a) / 0.1

4 0

3 years ago

Phoebe wants to get atleast a grade of 90 in the math class. Her grade will be the average of two tests. She scored a 94 on the

marin [14]

First things first, you will have 94 + x. Place those on top of a fraction bar and on the bottom put the two numbers over 2. So you have 94 + x over 2. Then, write the symbol ‘greater than or equal to’ and on the other side place your 90. So, 90 + x over 2 is greater than or equal to 90. This is a difficult thing to write out and explain so I hope this makes sense. :)

7 0

3 years ago

Read 2 more answers

1. M is the midpoint of LN and O is the midpoint of NP.

Reptile [31]

1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP

If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8

2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2

Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4

FC= 2/(1+2) * AC
FC= 2/3 * 12= 8

3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2

DC=3x+12= 3(-2) +12= 6

4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3

C
BC lines slope would be (-1)-(-1)/4 = 0/4.
The line would be
0=x+a
a=-1
0=x-1
x=1

y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1

y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)

5.
a. Circumcenter: the intersection of perpendicular bisector lines<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines

5 0

3 years ago

The graphs below have the same shape. What is the equation of the blue graph?

mario62 [17]

Answer:

c) G(x) = (x + 1)**2

Step-by-step explanation:

A function left shift is x + shift and a right shift is x - shift.

5 0

3 years ago