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3 years ago

Explain what circumference and area have in common and how they are different.

Mathematics

2 answers:

Advocard [28]3 years ago

8 0

Answer:

Circumference is the distance around a circle, or the perimeter. Area is the space inside the circle. Area has square units, circumference does not. They both have π in the formula.

Step-by-step explanation:

cluponka [151]3 years ago

3 0

Circumference is normally used for circles and is asking for the outer edge of the circle while area is the inner part of the shape formulas for circumference is 2 x PIE x radius
While area is length x breadth for most shapes or PIE x (radius)squared so both have to do with measurements and differ in what the measurements are focusing on

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What is 0.5 + -1 1/2

lakkis [162]

-1 is your answer.........

6 0

3 years ago

How do you write 144% as a fraction, mixed number, or whole number in simplest form?

Dmitrij [34]

In fraction, 144/100

Mixed number = 1 44/100

Simplest form = 144

3 0

3 years ago

In a survey of 12 people, it was found that an average of $36 was spent on their child's last birthday gift with a standard devi

Nataliya [291]

Answer:

(27.3692 ; 44.6308)

Step-by-step explanation:

Mean, xbar = 36

Standard deviation, s = 11

Sample size, n = 12

Tcritical at 0.2, df = 12 - 1 = 11 ; Tcritical = 2.718

Confidence interval :

Xbar ± Margin of error

Margin of Error = Tcritical * s/sqrt(n)

Margin of Error = 2.718 * 11/sqrt(12) = 8.6308

Confidence interval :

Lower boundary : 36 - 8.6308 = 27.3692

Upper boundary : 36 + 8.6308 = 44.6308

(27.3692 ; 44.6308)

3 0

3 years ago

Please help 30 points Asap

Andrej [43]

40 units2˛

Looking at the figure, the rectangle has the vertexes (2,1), (3,-3), (-5,-5) and (-6,-1). The parallelogram has the vertexes (2,7), (3,3), (3,-3), and (2,1).

The area of a parallelogram is base times height. We have 2 vertical lines at x=2 and x=3, so the height is 1. And the length of the line from (3,3) to (3,-3) is 6, so the base is 6. Therefore the area of the parallelogram is 1*6 = 6.

The rectangle is a tad trickier since it's not aligned with either the x or y axis. But we can use the Pythagorean theorem to get the lengths.

L = sqrt((2 - -6)^2 + (1 - -1)^2)

L = sqrt(8^2 + 2^2)

L = sqrt(64 + 4)

L = sqrt(68) = 2*sqrt(17)

W = sqrt((2-3)^2 + (1- -3)^2)

W = sqrt((-1)^2 + 4^2)

W = sqrt(1 + 16)

W = sqrt(17)

And the area is length * width, so:

2*sqrt(17)*sqrt(17) = 2 * 17 = 34

And the total area is the sum of the areas, so

34 + 6 = 40

So the area of the figure is 40 square units.

7 0

3 years ago

Read 2 more answers

Mathematical induction, prove the following two statements are true

adelina 88 [10]

Prove:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}$
____________________________________________

Base Step: For n=1:
$n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1$
and
$4-\dfrac{n+2}{2^{n-1}}=4-3=1$
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}$
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}$

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
$=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}$

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
$1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}$

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
$=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}$

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
$=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}$

Distribute the -2 and combine the fractions together,
$=4+\dfrac{-2k-4+(k+1)}{2^{k}}$

Combine like-terms,
$=4+\dfrac{-k-3}{2^{k}}$

pull the negative back out,
$=4-\dfrac{k+3}{2^{k}}$

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
$=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}$

3 0

3 years ago