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zysi [14]
3 years ago
5

Find the range of the quadratic function. (y=2x^2-20x+53)

Mathematics
1 answer:
Anit [1.1K]3 years ago
4 0

Answer: The range is y ≥ 3.

Step-by-step explanation:

When we have an equation:

y = f(x)

The range is the set of the possible values of y.

In this case, we have:

y = f(x) = 2*x^2 - 20*x + 53

This is a quadratic equation, and the leading coefficient is positive. This means that the arms of the graph will open upwards.

Then the minimum value of y will be the vale at the vertex.

We know that for a quadratic equation of the form:

a*x^2 + b*x + c

The vertex is at:

x = -b/2a

Then in this case, the vertex will be at:

x = -(-20)/(2*2) = 20/4 = 5

Then the smallest value of this function will be:

f(5) = 2*5^2 - 20*5 + 53 = 3

This is the smallest value that y can take, and y can take any value greater than 3.

Then the range can be written as:

y ≥ 3.

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Answer this question please; number 5... show all work thank you
My name is Ann [436]

Answer:

rate of the plane in still air is 33 miles per hour and the rate of the wind is 11 miles per hour

Step-by-step explanation:

We will make a table of the trip there and back using the formula distance = rate x time

                 d              =            r       x       t

there

back

The distance there and back is 264 miles, so we can split that in half and put each half under d:

               d            =         r        x        t

there    132

back     132

It tells us that the trip there is with the wind and the trip back is against the wind:

           

              d           =        r         x        t

there     132        =    (r + w)

back      132       =     (r - w)

Finally, the trip there took 3 hours and the trip back took 6:

           

             d         =        r        *        t

there    132     =    (r + w)     *       3

back     132     =    (r - w)      *       6

There's the table.  Using the distance formula we have 2 equations that result from that info:

132 = 3(r + w)        and        132 = 6(r - w)

We are looking to solve for both r and w.  We have 2 equations with 2 unknowns, so we will solve the first equation for r, sub that value for r into the second equation to solve for w:

132 = 3r + 3w and

132 - 3w = 3r so

44 - w = r.  Subbing that into the second equation:

132 = 6(44 - w) - 6w and

132 = 264 - 6w - 6w and

-132 = -12w so

w = 11

Subbing w in to solve for r:

132 = 3r + 3(11) and

132 = 3r + 33 so

99 = 3r and

r = 33

7 0
3 years ago
Choose the best definition for the following term: variable
andrew11 [14]

Step-by-step explanation:

a variable is a quantity that may change within the context of a mathematical problem or experiment

I hope this was helpful

6 0
3 years ago
Please write the answer​
Travka [436]

\sf \dfrac{3^x - 5 \times 3^{(x-2)}}{3^{(x-3)}} \\ \\ \longrightarrow \sf \dfrac{ {3}^{x} }{ {3}^{(x - 3)}} - \frac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)}  \\ \\ \longrightarrow \sf {3}^{[x - (x - 3)]} - \dfrac{5}{ {3}^{(x - 3)} } \times {3}^{(x - 2)} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ {3}^{(x - 3)} } \times \dfrac{ {3}^{x}}{9} \\ \\ \longrightarrow \sf {3}^{3} - \dfrac{5}{ \bigg(\dfrac{ {3}^{x} }{ {3}^{3} }\bigg) } \times \dfrac{ {3}^{x}}{9}\\ \\ \longrightarrow \sf {3}^{3} - \dfrac{ ({3}^{3})( 5)}{ {3}^{x} } \times \dfrac{ {3}^{x}}{9}\\ \\ \sf \longrightarrow {3}^{3} - (5 \times 3) \\  \\ \longrightarrow \sf \: 27 - 15 \\  \\ \longrightarrow \leadsto{\underline{\boxed{\sf{ \pink{ 12}}}}}

6 0
2 years ago
Solve. Check the solution. t/8 = 15 *
sp2606 [1]

Answer:

the answer is 15 I'm positive

8 0
2 years ago
Read 2 more answers
50 people plan to attend the dance and each person will eat 2 slices of cake. If each cake contains 20 slices, how many cakes sh
Naya [18.7K]

Answer:

5

Step-by-step explanation:

You <em>multiply</em> the number of people by the number of each slices each person gets. Which is 100; 50x2.

Each cake has 20 slices so you <em>divide</em> 100 by 20. Which equals 5.

It would take 5 cakes to feed all 50 people 2 slices each.

<em>Hope this helps</em>. :)

3 0
3 years ago
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