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Mrrafil [7]
1 year ago
12

Over which interval(s) is the function decreasing?A) -4 < x < 3B) -0.5 < x < ∞C) -∞ < x < -0.5D) -∞ < x &lt

; -4

Mathematics
1 answer:
Novay_Z [31]1 year ago
5 0

In the interval where the function is decreasingcreasing, the input or x values increase as the output or y values decrease. Looking at the graph, moving from the left to the right, the values of x are increasing whie the values of y are decreasing. This trend continued till we got to x = 0.5. Thus, in the interval from negative infinity to x = - 0.5, the function was decreasing.

The correct option is C

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There are multiple chickens and rabbits in a cage. There are 72 heads and 200 feet inside of the cage. How many chickens are in
sleet_krkn [62]
Given:
Chicken
Rabbit
72 heads
200 feet

Chicken - 1 head , 2 feet
Rabbit - 1 head , 4 feet

C + R = 72
2C + 4R = 200

C = 72 - R
2(72-R) + 4R = 200
144 - 2R + 4R = 200
2R = 200 - 144
2R = 56
R = 56/2
R = 28

C = 72 - 28 
C = 44

2C + 4R = 200
2(44) + 4(28) = 200
88 + 112 = 200
200 = 200

There are 44 CHICKENS and 28 RABBITS.
3 0
3 years ago
Find the volume of the cylinder.
Nadya [2.5K]

Answer:

169

Step-by-step explanation:

V=π (x) r (x) 2 (x)

h= π·32·6

≈169.646


4 0
4 years ago
PLZZZZZZ HELPPPPP ASAP!!!!!!!!!!!!!!!!!!
Ksivusya [100]

Answer:

fraction 1 over 3 whole squared m2

Step-by-step explanation:

we know that

The area of a square is

A=b^{2}

where

b is the length side of a square

In this problem we have

b=\frac{1}{3}\ m

substitute

A=(\frac{1}{3})^{2}

A=\frac{1}{9}\ m^{2}

therefore

The expression that can be used to find the area of the square is

fraction 1 over 3 whole squared m2

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

<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
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Nonamiya [84]

Answer: uhhhhhhhhhhhhhhhhhhh whats the question?

Step-by-step explanation:

5 0
3 years ago
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