the perimeter of a pool table is 30 ft.The table is twice as long as it is wide.what is the length of the pool table.

Answers

Answer 1

Answer: perimeter is legnth +legnth + width + width or
P=2L+2W

L=2W
subsitute L=2W for L in P=2L+2W

P=2(2W)+2W
P=4W+2W
P=6W

we know that Perimiter=30 so
30=6W
divide both sides by 6
5=Width

subsitute W=5 for W in L=2W
L=2(5)
L=10

to check
10=2(10)+2(5)
30=20+10
30=30
checks

Legnght=10
Width=5

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5. Find the equation of the line through the point (1, 3) with a slope of 2 in both pointslope form and slope-intercept form. Sketch a graph of the function.

Answers

Answer:

point slope form: y-3=2(x-1)

Slope-intercept form: y=2x+1

Step-by-step explanation:

y-y=m(x-x)

y-3=2(x-1)

y-3=2x-2

y=2x-2+3

y=2x+1

The expression (x8 - 48) can be factored as (x2 - 42)2 (x2 + 42)2.
a. True
b. False

Answers

I think the answer is false, but I'm not sure. I wish I could help more ):

I am confused on this , I’ve tried twice and got it wrong.

Answers

Answer:

$f^-^1(x)=-x^2+6x-5 \ \text{for the domain}\ [3, \infty)$

Step-by-step explanation:

Consider the function $f(x)=√(4-x)+3$ for the domain $(- \infty, 4]$ .

Find $f^-^1(x)$ , where f^(-1) is the inverse of f.

Also state the domain of f^(-1) in interval notation.

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We can start solving this problem by finding the inverse of f(x). This is done by switching the x- and y- variables, and solving for y.

$y=√(4-x)+3 \rightarrow x=√(4-y)+3$
$x=√(4-y)+3$

We can start solving for y by subtracting 3 from both sides of the equation.

$x-3=√(4-y)$

Get rid of the radical by squaring both sides of the equation.

$(x-3)^2=(√(4-y))^2$
$(x-3)(x-3)=4-y$

Use FOIL to multiply the binomial (x-3) together.

$x^2-3x-3x+9=4-y$

Combine like terms.

$x^2-6x+9=4-y$

Subtract 4 from both sides of the equation.

$x^2-6x+5=-y$

Divide both sides of the equation by -1.

$-x^2+6x-5=y$
$f^-^1(x)=-x^2+6x-5$

---

The domain and range of a function are flipped for its inverse, meaning that to find the domain of the inverse function, you can find the range of the original function f(x), and that will be your inverse function's domain.

The range of $f(x)=√(4-x) +3$ is $y \geq 3$ , since the vertical shift of the graph is at k = 3. You can also graph this function on a calculator to see that the graph does indeed start at y = 3.

Now that we know the domain and range of the original function, we know that these are flipped for the inverse function.

Original function:

Domain: $x\leq 4$
Range: $y\geq 3$

Inverse function:

Domain: $x\geq 3$
Range: $y\leq 4$

The final answer is:

The inverse $f^-^1(x)=-x^2+6x-5 \ \text{for the domain}\ [3, \infty)$ .

You can also write the domain as: $x\geq 3$ .

Hi, may someone help me with this question? Thank you!:)“If f(x) = x^2 + 7, find f(x+2)”

Answers

Answer:

=x^2 +4x+11

Step-by-step explanation:

f(x) = x^2 + 7,

Replace x with x+2

f(x+2) = (x+2)^2 + 7

= (x+2)(x+2) +7

FOIL

= x^2 +2x+2x+4 +7

Combine like terms

=x^2 +4x+11

Answer:

$f(x+2)=x^2+4x+11$

Step-by-step explanation:

In $f(x)=x^2+7$ , for all values of $x$ , we substitute $x$ (what is in the parentheses) into $x^2+7$ to output a $y$ value.

In $f(x+2)$ , the term $(x+2)$ is in the parentheses. Therefore, substitute $(x+2)$ for $x$ in $x^2+7$ to find $f(x+2)$ :

$f(x+2)=(x+2)^2+7$

Expand using $(a+b)^2=a^2+2ab+b^2$ ,

$f(x+2)=x^2+4x+4+7$

Combine like terms:

$\boxed{f(x+2)=x^2+4x+11}$

Determine which ordered pair is a solution of y=8x.A. (0,8)
B. (-1,8)
C. (1.5,10)
D. (2,16)

Answers

we know that

If the ordered pair is a solution of the equation, then the ordered pair must satisfied the equation

we will proceed to solve each case to determine the solution of the problem

we have

$y=8x$ -------> equation $1$

case a) $(0,8)$

$x=0\ny=8$

For $x=0$

substitute the value of x in the equation $1$ and then compare the values of y

$y=8*0=0$

$0\neq 8$

The ordered pair case a) is not solution

case b) $(-1,8)$

$x=-1\ny=8$

For $x=-1$

substitute the value of x in the equation $1$ and then compare the values of y

$y=8*(-1)=-8$

$-8\neq 8$

The ordered pair case b) is not solution

case c) $(1.5,10)$

$x=1.5\ny=10$

For $x=1.5$

substitute the value of x in the equation $1$ and then compare the values of y

$y=8*1.5=12$

$12\neq 10$

The ordered pair case c) is not solution

case d) $(2,16)$

$x=2\ny=16$

For $x=2$

substitute the value of x in the equation $1$ and then compare the values of y

$y=8*2=16$

$16=16$

The ordered pair case d) is a solution

therefore

the answer is

$(2,16)$

y = 8x
y = 8(2)
y = 16
(x, y) = (2, 16)

The answer is D.

What will multiply to 16 and add to 1?

Answers

37 is the right answer because 47 is not mutable and 27 is too short

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