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Pani-rosa [81]
3 years ago
14

What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 27 - x2?

Mathematics
2 answers:
Likurg_2 [28]3 years ago
6 0

Answer:

108 units squared

Step-by-step explanation:

Length is 2x , width is y

area = 2x * y

sub for y

area = 2x(27 - x^2)

area = 54x - 2x^3

1st derivative: 54 - 6x^2

54 - 6x^2 = 0

6(9 - x^2) = 0

6(3 - x)(3 + x) = 0

so, x = 3

solve for y: 27 - 3^2 = 18

area = (3*2)* 18 = 108 units squared

statuscvo [17]3 years ago
5 0
The answer is 54 square units.
let the vertex in quadrant I be (x,y) 
<span>then the vertex in quadratnt II is (-x,y) </span>
<span>base of the rectangle = 2x </span>
<span>height of the rectangle = y </span>
<span>Area = xy </span>
<span>= x(27 - x</span>²<span>) </span>
<span>= -x</span>³<span> + 27x </span>
<span>d(area)/dx = 3x</span>²<span> - 27 </span><span>= 0 for a maximum of area </span>
<span>3x</span>²<span> = 3 x 3</span>² = <span>27 </span>
<span>x</span>²<span> = 9 </span>
<span>x = ±3 </span>
<span>y = 27-9 = 18 </span>
So, the largest area = 3 x 18 = 54 square units
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Convert the improper fraction into a mixed number:

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These numbers do not equal each other so this is a false.

68/5 - 22/5 = 9 1/5

Subtract the numerators on the left side of the equation:

46/5 = 9 1/5

Convert the improper fraction into a mixed number:

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Find the common difference for the sequence below. 1/4, 5/16, 3/8,..
vredina [299]

Answer:

the common difference of the sequence is (\frac{1}{16})

Step-by-step explanation:

A sequence of numbers is said to be in arithmetic progression of each term of the sequence has same common difference.

Now, here first term a1 = 1/4

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Now, a3 - a2 =  \frac{3}{8}  - \frac{5}{16}  = \frac{6 - 5}{16}  = \frac{1}{16}

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2 years ago
Read 2 more answers
A line tangent to the curve f(x)=1/(2^2x) at the point (a, f(a)) has a slope of -1. What is the x-intercept of this tangent?
kirza4 [7]

Answer:

x-intercept = 0.956

Step-by-step explanation:

You have the function f(x) given by:

f(x)=\frac{1}{2^{2x}}   (1)

Furthermore you have that at the point (a,f(a)) the tangent line to that point has a slope of -1.

You first derivative the function f(x):

\frac{df}{dx}=\frac{d}{dx}[\frac{1}{2^{2x}}]  (2)

To solve this derivative you use the following derivative formula:

\frac{d}{dx}b^u=b^ulnb\frac{du}{dx}

For the derivative in (2) you have that b=2 and u=2x. You use the last expression in (2) and you obtain:

\frac{d}{dx}[2^{-2x}]=2^{-2x}(ln2)(-2)

You equal the last result to the value of the slope of the tangent line, because the derivative of a function is also its slope.

-2(ln2)2^{-2x}=-1

Next, from the last equation you can calculate the value of "a", by doing x=a. Furhtermore, by applying properties of logarithms you obtain:

-2(ln2)2^{-2a}=-1 \\\\2^{2a}=2(ln2)=1.386\\\\log_22^{2a}=log_2(1.386)\\\\2a=\frac{log(1.386)}{log(2)}\\\\a=0.235

With this value you calculate f(a):

f(a)=\frac{1}{2^{2(0.235)}}=0.721

Next, you use the general equation of line:

y-y_o=m(x-x_o)

for xo = a = 0.235 and yo = f(a) = 0.721:

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The last is the equation of the tangent line at the point (a,f(a)).

Finally, to find the x-intercept you equal the function y to zero and calculate x:

0=-x+0.956\\\\x=0.956

hence, the x-intercept of the tangent line is 0.956

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