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leonid [27]
2 years ago
12

Genesis throws a ball up in the air. The graph below shows the height of the ball hh in meters after tt seconds. Find the interv

al for which the ball’s height is decreasing.
help please

Mathematics
1 answer:
Juli2301 [7.4K]2 years ago
5 0

We conclude that the interval in which the function decreases is (1.5, 3]

<h3>In which interval is the ball's height decreasing?</h3>

The ball is decreasing when the graph, reading from left to right, is going downwards.

Particularly, in this parabola that opens downwards, the graph will decrease after the vertex.

And we can see that the vertex is at x = 1.5

And we also see that the graph of the function ends at x = 3.

Then we conclude that the interval in which the function decreases is (1.5, 3]

If you want to learn more about parabolas:

brainly.com/question/4061870

#SPJ1

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Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 126 in. Find
Zarrin [17]

Answer:

The package of maximum dimensions has:

Width and height = 21 in each , and length = 42 in

Step-by-step explanation:

A package of square cross section has a girth equal to the perimeter of the square ("4 x" if we consider "x" as the side of the square). This quantity, added to the package's length "L" has to be no more than 126 in.

The maximum allowed for priority mail is therefore: 4x+L=126 \,in

At the same time, we want the volume of the package to be a maximum. The volume of this parcel is defined as the product of all three dimensions of the package:

V=width\,*\,height\,*\,length\\V=x\,*\,x\,*\,L\\V=x^2\,L

We can now use the first formula for the priority mail maximum dimensions, to write L in terms of "x" and replace it in the volume formula:

4x+L=126 \,in\\L=126-4x

So now the volume expression becomes:

V=x^2\,(126-4x)=126x^2-4\,x^3

If the student doesn't know calculus, a graphing tool can be used to find the maximum.

With calculus derivatives the maximum can be easily found:

We can request the derivative of the volume to satisfy the conditions for derivative = 0 and the function concave down to locate the function's "maximum".

V'(x)=252\,x-12\,x^2\\0=252\,x-12\,x^2\\0=12\,x\,(21-x)

This tells us that there are two solutions to the derivative equal zero:

x=0\,\,and\,\,x=21

The first solution is clearly a minimum since it would render a package of zero volume. the second solution (x=21) is the one that corresponds to the maximum of the volume function.

Therefore, the dimensions of the package of largest volume are: width and height: 21 in each, and length L= 126 - 4*21 = 42 in

8 0
3 years ago
Help me with question 8 Please, I will give a thanks
Sliva [168]
Typically 1211 snack mix would be left but the answer is 23. because it is 23

6 0
3 years ago
HELP ASAP ILL GIVE BRAINLIST
musickatia [10]

Answer:

TRUE If substituting the test point produces a false solution, we shade on the opposite side of the line.

FALSE Elimination will give you an exact answer to a system of equations.

Step-by-step explanation:

Weird, I thought I already answer this xD

8 0
3 years ago
PLS HELP ME WITH 28 ASAP!! (MUST SHOW WORK!!) + LOTS OF POINTS
Igoryamba
Multiply both by -1: 113.75-112=1.75
multiply answer by -1: -1.75
5 0
3 years ago
Read 2 more answers
Ort each equation according to whether it has one solution, infinitely many solutions, or no solution
NeX [460]
The first one is infinite many solutions
The second one is (x=0) one solution
The third one is (x=3) one solution
The fourth one is no solution
The last one is no solution
6 0
3 years ago
Read 2 more answers
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