The mass of a textbook is about 1.25 kilograms. About how many pounds is it?

The cost of insurance can vary wildly between insurance companies. True or False

Liono4ka [1.6K]

Answer:

The answer will be True

Step-by-step explanation:

Because I said so

4 0

3 years ago

Which expression can be simplified as

DedPeter [7]

Answer:

4th is correct answer. 1, 2, 3 not correct you can check the exponents properties

3 0

4 years ago

PLEASE HELP QUICKLY!!! There is a tree standing in the desert; the sun is rising to the east.

a_sh-v [17]

A. As sun peeks, it creates a straight line- sun to tree to shadow, angle=180°
b. During morning, the sun-tree-shadow changes from 180° to 90°, during noon, angle =90°.During sunset the angle changes from 90 to 0°.
c. Acute angles created in afternoon.
d. A right angle would be created at noon.
e. Obtuse angle would be created in morning.
f. Yes, a straight angle would be created at sunrise.
g. When sun is at mid-morning, the angle=45°+90°= 135°

4 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

$0.028x=0.68$

$x=24.3$

<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

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

Find x, and then any unknown angles in the quadrilateral.<br> x= ?

sertanlavr [38]

Answer:

x=65

other angle= 165

Step-by-step explanation:

quadrilaterals always have a total of 360 degrees

because of this you can set up an equation and solve

360=90+40+x+(4x-95)

360=130+5x-95

360=35+5x

325=5x

65=x

plug this in to find missing angle

4(65)-95=165

7 0

3 years ago