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kirza4 [7]
2 years ago
12

Draw at least six different sized rectangles that have an area of 64 square units

Mathematics
1 answer:
olga_2 [115]2 years ago
3 0

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

A=x*y

A=64\ units^{2}

so

x*y=64 --------> equation 1

let's assume different values of x to get the different values of y

<u>case 1)</u> For x=64 units

substitute in the equation 1

x*y=64

y=64/x

y=64/64

y=1\ units      

the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

<u>case 2)</u> For x=32 units

substitute in the equation 1

x*y=64

y=64/x

y=64/32

y=2\ units

the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

<u>case 3)</u> For x=16 units  

substitute in the equation 1

x*y=64

y=64/x

y=64/16

y=4\ units

the dimensions of the rectangle are 16 units x 4 units

see the draw in the attached figure N 3

<u>case 4)</u> For x=30 units

substitute in the equation 1

x*y=64      

y=64/x

y=64/30

y=\frac{32}{15} =2\frac{2}{15} \ units

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

<u>case 5)</u> For x=40 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/40

y=1.60\ units

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

<u>case 6)</u> For x=60 units

substitute in the equation 1

x*y=64    

y=64/x

y=64/60

y=\frac{16}{15} =1\frac{1}{15} \ units

the dimensions of the rectangle are 60 units x 1 (1/15) units

see the draw in the attached figure N 6

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Step-by-step explanation:

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lions [1.4K]

Answer:

b=9,a=12

Step-by-step explanation:

I solved by substitution:

b = 3/4a  ---> plug this into the other equation

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Now solve for b:

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Step-by-step explanation:

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I am having trouble with this relative minimum of this equation.<br>​
Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach.  You have to let me know if you want this done another way.

Here are some rules I'm going to use:

(f+g)'=f'+g'       (Sum rule)

(cf)'=c(f)'          (Constant multiple rule)

(x^n)'=nx^{n-1} (Power rule)

(c)'=0               (Constant rule)

(x)'=1                (Slope of y=x is 1)

y=4x^3+13x^2-12x-56

y'=(4x^3+13x^2-12x-56)'

y'=(4x^3)'+(13x^2)'-(12x)'-(56)'

y'=4(x^3)'+13(x^2)'-12(x)'-0

y'=4(3x^2)+13(2x^1)-12(1)

y'=12x^2+26x-12

Now we set y' equal to 0 and solve for the critical numbers.

12x^2+26x-12=0

Divide both sides by 2:

6x^2+13x-6=0

Compaer 6x^2+13x-6=0 to ax^2+bx+c=0 to determine the values for a=6,b=13,c=-6.

a=6

b=13

c=-6

We are going to use the quadratic formula to solve for our critical numbers, x.

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}

x=\frac{-13 \pm \sqrt{169+144}}{12}

x=\frac{-13 \pm \sqrt{313}}{12}

Let's separate the choices:

x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}

Let's approximate both of these:

x=0.3909838 \text{ or } -2.5576505.

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

f(x)=4x^3+13x^2-12x-56

f'(x)=12x^2+26x-12

f''(x)=24x+26

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here:  x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

24(-2.6)+25

-37.5  

So we have a maximum at x=-2.6.

24(.4)+25

9.6+25

34.6

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56

I'm shoving this into a calculator:

y=-58.4654411

So the approximate relative minimum is (0.4,-58.5).

If you graph y=4x^3+13x^2-12x-56 you should see the graph taking a dip at this point.

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Answer:

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Step-by-step explanation:

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