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2 years ago

You work 37 1/2 hours and earn $352.50. What is your hourly wage

Mathematics

1 answer:

umka21 [38]2 years ago

8 0

Answer:

$9.40

Step-by-step explanation:

well since you work 37.5 hours and you earn $352.50 you divide them to get your hourly wage

352.5 ÷ 37.5 = 9.4

so your hourly wage is $9.40

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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

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

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

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

$0 \leq x \leq 48.6\ ft$

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

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2 years ago

Draw to show 55 three different ways

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Needs more information.

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2 years ago

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Latrell bought 4 bags of powdered sugar. He got a total of 5 1/2 cups of sugar. How many cups of sugar were in each bag?

Stels [109]

So all 4 bags of sugar yielded 5 and 1/2 cups..... well.. how many in each one? well, let's split the 5 and 1/2 evenly in 4 pieces.

$\bf \stackrel{mixed}{5}\frac{1}{2}\implies \cfrac{5\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{11}{2}}\\\\
-------------------------------\\\\
\cfrac{\quad\frac{11}{2} \quad }{4}\implies \cfrac{\quad\frac{11}{2} \quad }{\frac{4}{1}}\implies \cfrac{11}{2}\cdot \cfrac{1}{4}\implies \cfrac{11}{8}\implies \stackrel{cups}{1\frac{3}{8}
}$

5 0

2 years ago

6 4/9 + 14 2/3= 5 1/4 – 3 3/5 =

dimulka [17.4K]

Refer to the attachments!!~