Could someone pleasee answer this?

Express 0.77230.77230, point, 7723 as a fraction.

geniusboy [140]

Answer:

your mom

Step-by-step explanation:

lol

6 0

3 years ago

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

Mnenie [13.5K]

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

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

so

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

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

$0.028x=0.68$

$x=24.3$

Alternative method

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

so

$0 \leq x \leq 48.6\ ft$

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

so

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

3 years ago

Solve 2(x + 1) = 2x + 2. A All real numbers B No solution C 0 D 1

FrozenT [24]

2(x+1)=2x+2 if you factor the right side ..

2(x+1)=2(x+1)

This is always true for any value of x so there are infinitely many solutions...

x= all real numbers.

6 0

3 years ago

Read 2 more answers

Solve the equation: 5/6 = 1/3 + D

Sauron [17]

Answer: $\frac{1}{2}$

Explanation:
$\frac{5}{6} = \frac{1}{3} + D$
$\frac{5}{6} - \frac{1}{3} = \frac{1}{3} - \frac{1}{3} + D$
$\frac{5}{6} - \frac{1}{3} = D$
$\frac{5}{6} - 2(\frac{1}{3}) = D$
$\frac{5}{6} - \frac{2}{6} = D$
$\frac{3}{6} = D$ = $\frac{1}{2}$

8 0

3 years ago

A contractor is given a scale drawing of a rectangular patio. The scale from the patio to the drawing is 4ft to 1in. What is the

AnnyKZ [126]

Answer:

170 ft²

Step-by-step explanation:

Given:

Patio drawing of 4.25 in by 2.5 in

Scale of Patio to its drawing = 4ft to 1 in

Requires:

Area of actual Patio

SOLUTION:

Area of actual Patio = area of a rectangle = length × width

Given a scale drawing of 4ft to 1 in, therefore:

Length of actual Patio = 4.25 × 4 = 17 ft

Width of actual Patio = 2.5 × 4 = 10 ft

Therefore:

Area of actual Patio = 17 ft × 10 ft = 170 ft²

5 0

3 years ago