Here is the problem: (Picture)

Computer A: A graph has programs on the x-axis and megabytes on the y-axis. A line goes through (0, 32) and (2, 176). Computer B

My name is Ann [436]

Answer:

its C

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Simplidy y square root 6 over 6 square root y

Contact [7]

Answer:

Step-by-step explanation:

You can multiply the fraction by 6 and then you are able to cancel out and find the simplification

4 0

3 years ago

HELP!!!!

ss7ja [257]

<h3>3 Answers: Choice D, Choice E, Choice F</h3>

============================================================

Explanation:

The inequality 6x - 10y ≥ 9 solves to y ≤ (3/5)x - 9/10 when you isolate y.

Graph the line y = (3/5)x - 9/10 and make this a solid line. The boundary line is solid due to the "or equal to" as part of the inequality sign. We shade below the boundary line because of the "less than" after we isolated for y.

Now graph all of the points given as I've done so in the diagram below. The points in the blue shaded region, or on the boundary line, are part of the solution set. Those points are D, E and F.

We can verify this algebraically. For instance, if we weren't sure point E was a solution or not, we would plug the coordinates into the inequality to get...

6x - 10y ≥ 9

6(5) - 10(2) ≥ 9 .... plug in (x,y) = (5,2)

30 - 20 ≥ 9

10 ≥ 9 ... this is a true statement

Since we end up with a true statement, this verifies point E is one of the solutions. I'll let you check points D and F.

-----------

I'll show an example of something that doesn't work. Let's pick on point A.

We'll plug in (x,y) = (-1,1)

6x - 10y ≥ 9

6(-1) - 10(1) ≥ 9

-6 - 10 ≥ 9

-16 ≥ 9

The last inequality is false because -16 is smaller than 9. So this shows point A is not a solution. Choices B and C are non-solutions for similar reasons.

3 0

3 years ago

We toss three coins. What is the probability of the event "3 Heads appear"?

My name is Ann [436]

1/8 why? because 1/2 for the first head. times 1/2 for the second head. times 1/2 for the third head

7 0

3 years ago

Read 2 more answers

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