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3 years ago

What type of number can be written as a fraction p%q where p and q aré integers and q is not equal to zero

1 answer:

7 0

Hi there!

The correct answer is Rational Number.

$\bf{EXPLANATION}$ :-

Rational Number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator " p " and a non-zero denominator " q ".

~ Hope it helps!

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X - y= 15 4x+ 2y= 30

Ivahew [28]

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3 years ago

What is two thirds to the negative cubed power¿

Mashutka [201]

(23)−3
(32)3
(32)*(32)*(32)
3*3*32*2*2=3323
278(Decimal: 3.375)

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3 years ago

Which ordered pair is a solution to the inequality? Help please!!

Murrr4er [49]

Choice A is the answer which is the point (1,-1)

=========================================

How I got this answer:

Plug each point into the inequality. If you get a true statement after simplifying, then that point is in the solution set and therefore a solution. Otherwise, it's not a solution.

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checking choice A

plug in (x,y) = (1,-1)

$y - 2x \le -3$

$-1 - 2(1) \le -3$

$-3 \le -3$

This is true because -3 is equal to itself. So this is the answer.

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

checking choice B

plug in (x,y) = (2,4)

$y - 2x \le -3$

$4 - 2(2) \le -3$

$0 \le -3$

This is false because 0 is not to the left of -3, nor is 0 equal to -3. We can cross this off the list.

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

checking choice C

plug in (x,y) = (-2,3)

$y - 2x \le -3$

$3 - 2(-2) \le -3$

$7 \le -3$

This is false because 7 is not to the left of -3, nor is 7 equal to -3. We can cross this off the list.

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

checking choice D

plug in (x,y) = (3,4)

$y - 2x \le -3$

$4 - 2(3) \le -3$

$-2 \le -3$

This is false because -2 is not to the left of -3, nor is -2 equal to -3. We can cross this off the list.

4 0

3 years ago

Question 6 of 25 Evaluate 4(x - 3) + 5x - x2 for x = 3. -

Zanzabum

$\text{Given that,}\\\\4(x-3) +5x -x^2\\\\\text{For x =3,}\\\\4(3-3) +5(3) -3^2 = 4(0) +15 -9 = 6$

6 0

3 years ago

Read 2 more answers

F(x)= x^2-8x+15

mash [69]

If I'm reading your equations correctly, they are:f(x)=x2-8x+15g(x)=x-3h(x)=f(x)/g(x)The domain of a function is the set of all possible inputs, what we can plug in for our variable.The largest two limitations on domains (other than explicit limitations, like in piecewise functions) are radicals and rational functions. With radical expressions we know that we CANNOT take an even root of a negative number. I don't see that problem here. With rationals we know that we CANNOT divide by zero. So the question becomes, when does h(x) ask us to divide by zero? When is the denominator of h(x) zero?Since the denominator of h(x) is g(x), we cannot let g(x) equal zero. So when does that happen? when x-3=0 or when x=3. I hope you see here that if x=3, then g(x)=0, and so h(x)=f(x)/0, which we CANNOT do. The domain of h(x) is all real numbers not equal to 3. There is more going on here. If you had factored f(x) first, you could have written h(x) in a confusing way:h(x)=( f(x) ) / ( g(x) )h(x)= ( (x-5)(x-3) ) / (x-3) Right here, it looks like (x-3) will cancel out from the top and bottom of your fraction. It does, in a way. The graph of h(x) will behave exactly like the line y=x-5, except that it has a hole in it at x=3 (check this! it's cool!)SOOO, the takeaway is that it is better to determine limitations on your domain BEFORE over-simplifying your equations.

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3 years ago