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3 years ago

20 POINTS

Mathematics

1 answer:

Mashutka [201]3 years ago

6 0

To find x:

ON is parallel to LM

8x - 8 = 7x + 4

-7x -7x

1x - 8 = 4

+8 +8

1x = 12

--- ---

1 1

x = 12

8(12) - 8 = 7(12) + 4

96 - 8 = 84 + 4

88 = 88

------------------------------------

To find y:

NM is parallel to OL

x - 5 = 3y - 6

12 - 5 = 3y - 6

7 = 3y - 6

+6 +6

13 = 3y

--- ---

3 3

13/3 = y

Check:

12 - 5 = 3(13/3) - 6

7 = 13 - 6

7 = 7

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3 years ago

The students in Mr. Wilson's Physics class are making golf ball catapults. The

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Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

2 years ago

Please help me ! Reduced to lowest terms

Mandarinka [93]

Answer:

Step-by-step explanation:

7) 2ab + 4ac = 2*a*b + 2 *2 *a*c

= 2a(b + 2c)

$\dfrac{2ab^{2}c}{2ab+4ac}=\dfrac{2*a* b^{2}*c}{2a(b+2c)}\\\\\\\text{2a in the denominator and numerator get cancelled}\\\\ = \dfrac{b^{2}c}{b+2c}$

8) ac - a²c² = c( a - a²c)

bc - abc = c(b - ab)

$\dfrac{ac - a^{2}c^{2}}{bc-abc^{2}}=\dfrac{c(a-a^{2}c)}{c(b-ab)}\\\\$

$=\dfrac{a-a^{2}c}{b-abc}$

9) b²+ 4b - 5 = b² + 5b - 1b - 5

= b(b + 5) - 1(b + 5)

= (b +5)(b - 1)

b² + 8b + 15 = b² + 5b + 3b + 15

= b(b + 5) + 3(b + 5)

= (b + 5 )(b + 3)

$\dfrac{b^{2}+4b-5}{b^{2}+8b+15}=\dfrac{(b+5)(b-1)}{(b+5)(b+3)}\\\\\\$

$=\dfrac{b-1}{b+3}$

8 0

2 years ago