Ask question

Ask question

All categories

4 years ago

7

Danica has laid so they form a rectangle with a perimeter of 18 inches.What is the difference between the greatest and least pos

sible areas of the rectangle?

1 answer:

stealth61 [152]4 years ago

8 0

Elemantary school so
area=legnth times width
to get greatest area, make legnth as close to width as possible in measure

to get least area, make legnth as far away from width as possible in measure
perimiter=2 times legnth+2 times width=2 times (legnth+width)
18=2 times (legnth+width)
divide 2
9=legnth+width
find closest and farthest
1+9 is farthest
4+5 is closest

areas=legnth times with
1 times 9=9
4 times 5=20
differnce is subtract
20-9=11

answe is 11
(note I didn't use decimals since the user is supposedly in elemenatry school)

You might be interested in

During his workouts last week, Allen jogged the same amount of time each day for 6 days. Use the following information to determ

Nataly [62]

He rode his bike 45 minutes each day:
45x6 = 270 total time cycled

Jogging time = total workout time - cycle time/6

360-270= 90/6= 15 minutes jogged each day

8 0

3 years ago

Find the equation of the axis of symmetry for the parabola y = x^2 .

Tamiku [17]

The vertex of the parabola is <span><span>(<span>2,1</span>)</span><span>(<span>2,1</span>)</span></span> .

So, the axis of symmetry is the line <span><span>x=2</span><span>x=2</span></span> .

6 0

3 years ago

Read 2 more answers

Directions: What does each of the following mean? Write out in words and phrases.

GalinKa [24]

Eleven plus
t minus seventeen
ten multiplied by n
fifteen multiplied by y
four multiplied by x
a plus b
fourteen times x plus three times y

8 0

3 years ago

Select the correct answer.<br> What is the range of this absolute value function?

Oliga [24]

Answer:

I believe its c

Step-by-step explanation:

6 0

2 years ago

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Keith_Richards [23]

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step explanation:

Given

$r = $\sqrt[3]{2} + \sqrt[3]{4}$$ --- the root

Solving (a): The polynomial

A cubic function is represented as:

$f = (a + b)^3$

Expand

$f = a^3 + 3a^2b + 3ab^2 + b^3$

Rewrite as:

$f = a^3 + 3ab(a + b) + b^3$

The root is represented as:

$r=a+b$

By comparison:

$a = $\sqrt[3]{2}$

$b = \sqrt[3]{4}$$

So, we have:

$f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3$

Expand

$f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

Evaluate like terms

$f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)$

Recall that: $r = $\sqrt[3]{2} + \sqrt[3]{4}$$

So, we have:

$f = 6 + 6r$

Equate to 0

$f - 6 - 6r = 0$

Rewrite as:

$f - 6r - 6 = 0$

Express as a cubic function

$x^3 - 6x - 6 = 0$

Hence, the cubic polynomial is:

$f(x) = x^3 - 6x - 6$

Solving (b): Prove that r is irrational

The constant term of $x^3 - 6x - 6 = 0$ is -6

The divisors of -6 are: -6,-3,-2,-1,1,2,3,6

Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values

$f(-6) = (-6)^3 - 6*-6 - 6 = -186$

$f(-3) = (-3)^3 - 6*-3 - 6 = -15$

$f(-2) = (-2)^3 - 6*-2 - 6 = -2$

$f(-1) = (-1)^3 - 6*-1 - 6 = -1$

$f(1) = (1)^3 - 6*1 - 6 = -11$

$f(2) = (2)^3 - 6*2 - 6 = -10$

$f(3) = (3)^3 - 6*3 - 6 = 3$

$f(6) = (6)^3 - 6*6 - 6 = 174$

For r to be rational;

The divisors of -6 must divide f(x) without remainder

i.e. Any of the above values must equal 0

<em>Since none equals 0, then r is irrational</em>

3 0

3 years ago