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3 years ago

6

Write an expression to find the difference in the maximum area and minimum area of nys high school soccer field. Then evaluate y

ou expression.

1 answer:

Vlad1618 [11]3 years ago

6 0

You didn't give the numbers but an expression is a problem with no equals so you do t pit the total if that helps at all

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REALLY NEED HELP <br> What is the slope of the given line??

Montano1993 [528]

Answer:

3

Step-by-step explanation:

slope is rise over run

6/2 simplifyed is 3/1 or just 3

6 0

3 years ago

Boris chooses three different numbers the sum of the three numbers is 36 one of the numbers is a cube number the other two numbe

OlgaM077 [116]

Answer:

3,4,5

Step-by-step explanation:

Here, we need to asume that the values are integers, as if not, there will be infinite solutions.

Lets go by parts:

The 1st number is a cubic, lets call it A

The second is a factor of 20, lets call it B. The same for the 3rd, call it C.

So, A + B+ C = 36

There are no many factors of 20, those are the numbers that when multiplied gives as the value of 20. Those are 1, 2, 4, 5, 10 and 20. So, B and C are some of these numbers.

Then, we know A if less than 36, other way the whole sum will be greater than 36. How many cubic number with integer cubic roots are less than 36? Well, lets guess:

1^3 = 1 -> valid, as it is less than 36

2^3 = 8 -> valid, as it is less than 36

3^2 = 27 -> valid, less than 36

4^3 = 64 -> not valid, as it is greater than 36

So, A ir 1, 2 or 3.

If A is 1, then B+C needs to be 35. But, from the factors of 20 that we listed, there are no combinations of 2 numbers that sum 35. So, A CAN'T be 1.

If A = 2, then we have:

8 + B + C = 36

B + C = 28

And again, there are not combinations of two factors of 20 that sum 28 (try yourself).

If A=3:

27 + B + C = 36

B +C = 9

And we have a winner!!! If B=4 and C=5 (or viceversa C=4 and B=5)

27 + 4 + 5 = 36

So, A=3, B=4 and C=5 or A=3, C=4 and B=5 are solutions.

3 0

4 years ago

Which of the next equations are linear:

insens350 [35]

The first and third ones are both linear. The second one is not because of the exponent. Instead it is a parabola.

8 0

4 years ago

All integers are rational numbers true or false please explain

baherus [9]

True, integers are all counting numbers and their opposites, and all of them can be represented as a fraction
example: 3 is a counting number(integer) and it is rational because it can be represented as 3over1

8 0

4 years ago

The point (2, 1) is a solution to which of the following systems of equations?

Snezhnost [94]

$\huge\boxed{\left \{ {{x+y=3} \atop {3x-y=5}} \right.}$

We can solve this problem by testing the solution against each system of equations. We'll do this by substituting the point into each equation for each system.

<h2>System 1 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}2x+y&=5\\2(2)+1&=5\\4+1&=5\\5&=5\\&\checkmark\end{aligned}&\begin{aligned}-2x+y&=2\\-2(2)+1&=2\\-4+1&=2\\-3&=2\\&\text{X}\end{aligned}\end{array}$

<h2>System 2 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-2x-y&=-5\\-2(2)-1&=-5\\-4-1&=-5\\-5&=-5\\&\checkmark\end{aligned}&\begin{aligned}2x-y&=2\\2(2)-1&=2\\4-1&=2\\3&=2\\&\text{X}\end{aligned}\end{array}$

<h2>System 3 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-x+y&=3\\-2+1&=3\\-1&=3\\&\text{X}\end{aligned}&\begin{aligned}-3x+y&=-5\\-3(2)+1&=-5\\-6+1&=-5\\-5&=-5\\&\checkmark\end{aligned}\end{array}$

<h2>System 4 - $\Large\checkmark$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}x+y&=3\\2+1&=3\\3&=3\\&\checkmark\end{aligned}&\begin{aligned}3x-y&=5\\3(2)-1&=5\\6-1&=5\\5&=5\\&\checkmark\end{aligned}\end{array}$

Since both equations in the final system are true when solved with $(2, 1)$ , the answer is the last system.

3 0

3 years ago