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2 years ago

A square has sides that measure 6 inches each. What is the ratio of the perimeter of the square to the area of the square?

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

5 0

Answer:

2 : 3

Step-by-step explanation:

Given the square has a side of 6 in , then

perimeter = 4 × 6 = 24 in

area = 6² = 36 in²

Then ratio of perimeter : area

= 24 : 36 ( divide both parts by 12 )

= 2 : 3

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-2x-y-2z+4x-2y+3z Please help

sdas [7]

Answer:

2x-3y+z

Step-by-step explanation:

-2x-y-2z+4x-2y+3z

combine like terms

-2x+4x= 2x

-y-2y= -3y

-2z+3z= z

so..

2x-3y+z

8 0

3 years ago

Name two integers that have a product between -10 and -15

umka2103 [35]

A product means when they are multiplied together.
I.e. The product of 3 and 4 is 12, because 3 * 7 = 4
i.e. The product of -4 and + 8 is 4, because -4 * 8 = -36

An integer means a whole number like 1, 2, 3, 4, 5, or -1, -2, -3, -4.
i.e. 3.2 is NOT AN INTEGER since there is a decimal

Two integers that have a product between -10 and -15
> Can be -6 and +2 which the product is -12 <------- (-6 * +2)
> Can be -7 and +2 which the product is -14 <------- (-7 * -21)
So an answer would be
"-6 and +2" are integers which have a product between -10 and -15

6 0

3 years ago

You roll a twelve-sided die (having values one through twelve on its faces). What is the probability that the value of the roll

Tom [10]

I think it’s 1/12 exactly you have a one in 12 chances of rolling a 5

5 0

3 years ago

Read 2 more answers

What is 1 1/3 as a improper fraction

Mumz [18]

Okay so for this all you have to do is add 1 to 1/3.

However, to do this you first need to convert 1 into a fraction.

1 = 3/3
3/3 + 1/3 = 4/3

Hope this helps :)

8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

2 years ago