What is 5 + (9)(6)????????????

The area of a rectangular picture frame is 93.5 square inches. The area formula for a rectangle is A=l⋅w. Write and solve an equ

Nikolay [14]

Answer:

W=A/L

Step-by-step explanation:

If the area of the rectangle is L*W, you can just rearrange the equation A=L*W to get W=A/L, an equivalent equation.

5 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

Graph the system below and write its solution. 2X+Y=-6 Y= -1/3X-1

puteri [66]

2x + y = -6
y = ⁻¹/₃x - 1

2x + (⁻¹/₃x - 1) = -6
2x - ¹/₃x - 1 = -6
1²/₃x - 1 = -6
+ 1 + 1
1²/₃x = -5
1²/₃ 1²/₃
x = -3

y = ⁻¹/₃x - 1
y = ⁻¹/₃(-3) - 1
y = 1 - 1
y = 0
(x, y)= (-3, 0)

4 0

3 years ago

Is this increasing decreasing or constant F(x)=3x

Zanzabum

Hello :
this increasing function F(x)=3x because : the slope 3
is positif

7 0

3 years ago

X=(a+b)/2 solve for a

Usimov [2.4K]

Answer:

a=2x-b

Step-by-step explanation:

$x=\frac{a+b}{2}\\\\\frac{a+b}{2} = x\\ a+b = 2x\\a= 2x-b\\$

5 0

3 years ago