Calculate the present value. (Round your answer to two decimal places.)

The area formula for a rectangle = Length x Width
Let length = L and width = W.
We also know that the length is 3 yards longer than twice it's width.
SO this means L = 3 PLUS 2 times W
L = 3 + 2W
Now plug that into the area formula and we get:
Area = (3 + 2W) x W
You'll need to multiply everything in the brackets by W.
So what would (3 + 2W)x W = ?

4 0

3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Y = (4/9) x

frez [133]

Answer:

V = 8.06 cubed units

Step-by-step explanation:

You have the following curves:

$y_1=\frac{4}{9}x^2=f(x)\\\\y_2=\frac{13}{9}-x^2=g(x)$

In order to calculate the solid of revolution bounded by the previous curves and the x axis, you use the following formula:

$V=\pi \int_a^b [(g(x))^2-(f(x))^2]dx$ (1)

To determine the limits of the integral you equal both curves f=g and solve for x:

$f(x)=g(x)\\\\\frac{4}{9}x^2=\frac{13}{9}-x^2\\\\\frac{4}{9}x^2+x^2=\frac{13}{9}\\\\\frac{13}{9}x^2=\frac{13}{9}\\\\x=\pm 1$

Then, the limits are a = -1 and b = 1

You replace f(x), g(x), a and b in the equation (1):

$V=\pi \int_{-1}^{1}[(\frac{13}{9}-x^2)^2-(\frac{4}{9}x^2)^2]dx\\\\V=\pi \int_{-1}^1[\frac{169}{81}-\frac{26}{9}x^2+x^4-\frac{16}{81}x^4]dx\\\\V=\pi \int_{-1}^1 [\frac{169}{81}-\frac{26}{9}x^2+\frac{65}{81}x^4]dx\\\\V=\pi [\frac{169}{81}x-\frac{26}{27}x^3+\frac{65}{405}x^5]_{-1}^1\\\\V\approx8.06\ cubed\ units$

The volume of the solid of revolution is approximately 8.06 cubed units

8 0

4 years ago