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2 years ago

Which rectangles have the same areas but greater perimeters than the one shown below? Choose all that apply.

1 answer:

5 0

The rectangles have the same areas but greater perimeters are 12 ft by 3 ft, 9 ft by 4 ft and 18 ft by 2 ft.

Area is the amount of space occupied by a two dimensional shape or object.

A rectangle is shown with a length and height of 6 ft. Hence:

Area = (6 * 6) = 36 ft²

Perimeter = 2(6 + 6) = 24 ft

For 12 ft by 3 ft:

Area = (12 * 3) = 36 ft²

Perimeter = 2(12 + 3) = 30 ft

For 9 ft by 4 ft:

Area = (9 * 4) = 36 ft²

Perimeter = 2(9 + 4) = 26 ft

For 8 ft by 3 ft:

Area = (8 * 3) = 24 ft²

Perimeter = 2(8 + 3) = 24 ft

For 7 ft by 4 ft:

Area = (7 * 4) = 28 ft²

Perimeter = 2(7 + 4) = 26 ft

For 18 ft by 2 ft:

Area = (18 * 2) = 36 ft²

Perimeter = 2(18 + 2) = 40 ft

The rectangles have the same areas but greater perimeters are 12 ft by 3 ft, 9 ft by 4 ft and 18 ft by 2 ft.

Find out more on area at: brainly.com/question/25292087

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Prove that the Greatest Integer Function f: R -> R given by f(x) = [x], is neither one-once nor onto, where [x] denotes the g

Karo-lina-s [1.5K]

F: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
So, f(1.2) = f(1.9), but 1.2 ≠ 1.9
f is not one-one

Now, consider 0.7 ε R
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ε R such that f(x) = 0.7

So, f is not onto
Hence, the greatest integer function is neither one-one nor onto.

The answer was quite complicated but I hope it will help you.

8 0

4 years ago

Find a power series for the function, centered at c. f(x) = 1 9 − x , c = 4 f(x) = [infinity] n = 0 Incorrect: Your answer is in

Wewaii [24]

Looks like the given function is

$f(x)=\dfrac1{9-x}$

Recall that for |x| < 1, we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

We want the series to be centered around $x=4$ , so first we rearrange f(x) :

$\dfrac1{9-x}=\dfrac1{5-(x-4)}=\dfrac15\dfrac1{1-\frac{x-4}5}$

Then

$\dfrac1{9-x}=\displaystyle\frac15\sum_{n=0}^\infty\left(\frac{x-4}5\right)^n$

which converges for |(x - 4)/5| < 1, or -1 < x < 9.

4 0

4 years ago

What is equivalent to finding the “zeros” of a function?

Naddik [55]

Answer:

x-intercepts, solutions, roots.

Those are all I can think of. Hope that helps :)

5 0

3 years ago

1+4=5 2+5=7 3+6=21 8+11=

Karolina [17]

This is a common math problem. There are two ways to answer this question but both will give you the same answer. There is a misconception that this can also equal 40. If someone posts that the answer is 40, please let me know and I will explain why that is not actually the answer.

1+4=5 2+5=7
3+6=21
8+11=

These equations are actually:

1*(4+1)=52*(5+1)=7
3*(6+1)=21
8*(11+1)=96

Therefore:
8+11 = 96

3 0

4 years ago

Read 2 more answers

Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

Pachacha [2.7K]

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

5 0

4 years ago