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3 years ago

Please help with this question!!

Mathematics

1 answer:

xxMikexx [17]3 years ago

8 0

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}$

$x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}$

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If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

Step 1: Solve using the first point

(2, 28)

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

Step 2: Solve using the second point

(-1, -5)

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

Step 3: Solve using the third point

(4, 220)

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

Step 4: Solve using the fourth point

(-2, -20)

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

Step 5: Combine the first and fourth equations

$28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

Step 6: Solve for c in the second equation

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

Step 7: Substitute d with the stuff we got in step 5

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

Step 8: Substitute d and c into the first equation

$28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

Step 9: Substitute a, b, and c into the third equation

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

Step 10: Find a using b = 2

$a = b + 1$

$a = (2) + 1$

$a = 3$

Step 11: Find c using a = 3 and b = 2

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

Step 12: Find d using b = 2

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

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3 years ago

Read 2 more answers

Which of the following inequalities matches the graph? PLZ HELP DUE TODAY

balandron [24]

Answer:

A. 2x + y < 7

Step-by-step explanation:

the slope is negative and the y-intercept is positive

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2 years ago

Solving the whole thing

erik [133]

Let
x---------------> distance from people living to the city center

we Know that
Zone 1 covers people living within three miles of the city center
Zone 1 ------------> [x < 3 miles]

Zone 2 covers those between three and seven miles from the center
Zone 2 ------------> [ 3 <= x < = 7 miles]

Zone 3 covers those over seven miles from the center
Zone 3 ------------> [ x > 7 miles]

calculate the distance between two points to find the value of x

case A) point (0,0) point (3,4)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(4-0)² +(3-0)²]------> √[16+9]
x=√25-------------> x=5 miles

the answer Part A)
people living in (3,4)
x=5 miles -------------> covers Zone 2 [ 3 < =x <= 7 miles]

case B) point (0,0) point (6,5)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(5-0)² +(6-0)²]------> √[25+36]
x=√61-------------> x=7.81 miles

the answer Part B)
people living in (6,5)
x=7.81 miles -------------> covers Zone 3 [ x > 7 miles]

case C) point (0,0) point (1,2)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(2-0)² +(1-0)²]------> √[4+1]
x=√5-------------> x=2.23 miles

the answer Part C)
people living in (1,2)
x=2.23 miles -------------> covers Zone 1 [ x < 3 miles]

case D) point (0,0) point (0,3)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(3-0)² +(0-0)²]------> √[9]
x=√9-------------> x=3 miles

the answer Part D)
people living in (0,3)
x=3 miles -------------> covers Zone 2 [ 3 < =x <= 7 miles]

case E) point (0,0) point (1,6)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(6-0)² +(1-0)²]------> √[36+1]
x=√37-------------> x=6.08 miles

the answer Part E)
people living in (1,6)
x=6.08 miles -------------> covers Zone 2 [ 3 < = x <= 7 miles]

4 0

3 years ago