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Angelina_Jolie [31]
3 years ago
13

Amelia wants to know which school has the higher sat average relative to the resources invested per student and new your thought

of two different ways to Define this quantity identify these two definitions among the following options ​

Mathematics
1 answer:
Neko [114]3 years ago
5 0

Answer:

i don't know

Step-by-step explanation:

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if you have budgeted 12% of your monthly income for groceries how much can you spend on groceries if you earn $950 bi-weekly
finlep [7]
Well if you make 950 dollars every two weeks, in a month you will have 1,900. Then take 12% of 1,900, and you will have 228 dollars to spend a month on groceries.
7 0
3 years ago
Read 2 more answers
Does anyone have a coursehero account that I can borrow
vagabundo [1.1K]

Answer:

No but an easy way is to just make up random email names and passwords and on each acc u get 5 free questions,

Step-by-step explanation:

Can u pls mark me brainliest

5 0
3 years ago
Drag the expressions into the boxes to correctly complete the table.
lora16 [44]

Answer:

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form ax^n.

Here:

n = non-negative integer

a = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

x^4+\frac{5}{x^3}-\sqrt{x}+8

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains \sqrt{x}, so it is not a polynomial.

Also it contains the term \frac{5}{x^3} which can be written as 5x^{-3}, meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression x^4+\frac{5}{x^3}-\sqrt{x}+8 is not a polynomial.

Given the expression

-x^5+7x-\frac{1}{2}x^2+9

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

\left|x\right|^2+4\sqrt{x}-2

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

x^3-4x-3

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

\frac{4}{x^2-4x+3}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

3 0
3 years ago
Solve the system Also plz show work and what you did and these two equations with x and y the x and y have to be the same in bot
Gala2k [10]
Multply both equations by 2 leaving

x + 2y = 8
x - 2y = -2

Add the equations giving

2x = 6
x = 3

Substitute 3 in the top equation

3 + 2y = 8
2y = -5
y = -2.5
5 0
3 years ago
Need Help! I don't get it
serious [3.7K]
Well hmmm let's say you take the car and go in the city for 60 miles with it, well, the car can do 60 miles per gallon, since you just drove it for 60 miles, you only spent 1 gallon of gasoline then.

that only happens if you drive it for 60 miles, what if you drive it for more, let's do a quick table on that,

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
60&3.6(1)\\\\
&3.6\left( \frac{60}{60} \right)\\\\
120&3.6(2)\\\\
&3.6\left( \frac{120}{60} \right)\\\\
180&3.6(3)\\\\
&3.6\left( \frac{180}{60} \right)
\end{array}

and so on, now let's check if you less than 60 miles,

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
40&3.6\left( \frac{40}{60} \right)\\\\
20&3.6\left( \frac{20}{60} \right)\\\\
10&3.6\left( \frac{10}{60} \right)
\end{array}

so, if you divide the amount of miles driven, by 60, when you have driven it for 120 miles, 120/60 is just 2, and the cost is for 2 gallons, or 3.6 * 2, which is 7.2 bucks, for 180 miles is 180/60 or 3 gallons for 3.6 * 3 bucks, and so on.

now, what if you drive it instead for "m" miles?

\bf \begin{array}{ccll}
miles&cost\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
m&3.6\left( \frac{m}{60} \right)\\\\
\end{array}\implies c=3.6\left( \cfrac{m}{60} \right)\implies c=\cfrac{3.6m}{60}
\\\\\\
c=\cfrac{3.6}{60}m\implies c=0.06m
3 0
3 years ago
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