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3 years ago

Write the solution to the given inequality in interval notation.

Mathematics

2 answers:

Effectus [21]3 years ago

7 0

Is the last one
(-Infinity, -3] U (-1, positive infinity)

lozanna [386]3 years ago

5 0

(-infinity, -3] U (-1, infinity)
The last answer is correct

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What value should go in the empty box to complete the calculation for finding the product 73.458 × 0.39?

mr Goodwill [35]

Answer:

28.64862

Step-by-step explanation:

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3 years ago

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Julio practiced his trumpet on each 6 days during a week. What is the mean of minutes Julio practiced over these 6 days.

Novay_Z [31]

21+30+39+43+58+67 = 258.
258/6 = 43

43 minutes

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3 years ago

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PLEASE HELP ASAP 25 POINTS

Lisa [10]

We're going to go ahead and eliminate one answer out of the four. It cannot be B (the second answer) because the graph is made of dotted lines.

Dotted Lines = > or <

Solid Lines = ≤ or ≥

Next, let's focus on the straight line on the graph.

The equation for the line is

y = x - 4

Since the shaded region is below the line,

the equation will be y < x - 4

We can now eliminate answer A

Since the second equation is y < - l x - 2 l

It will be shaded below the graph since it uses the ( < ) (less than symbol)

This means the answer is C

Hope I helped, message me if you have any questions : )

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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

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

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

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

Kiran and his cousin work during the summer for a landscaping company. Kiran’s cousin has been working for the company longer, s

Fofino [41]

Sry I have no idea 768373

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3 years ago