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ExtremeBDS [4]
3 years ago
5

Need assistance with this one please

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
7 0

Answer:

(B) 4x^2y^2 + 5x + xy

Step-by-step explanation:

\dfrac{24x^3y^3 + 30x^2y + 6x^2y^2}{6xy}

6xy is a common factor.

= \dfrac{6xy(4x^2y^2 + 5x + xy)}{6xy}

6xy from numerator and denominator are canceled off.

= 4x^2y^2 + 5x + xy

zvonat [6]3 years ago
6 0

Answer:

C, 4x²y² + 5x + 1

Step-by-step explanation:

to solve, we can first factor the numerator by finding a GCF.

\frac{24x^3y^3+30x^2y+6xy}{6xy}

the GCF of the numerator would be 6xy, as each term can be divided by 6xy. we have the following:

6xy(\frac{4x^2y^2+5x+1}{6xy})

we can cancel out the numerator and the denominator as both contain 6xy so we have the following:

4x²y² + 5x + 1

we cannot simplify this any further, so our answer is C

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Any points that are on the bolded vertical line are on the y-axis, and any points on the bolded horizontal line is on the x-axis.

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Why do we need to learn Positive and Negative Integers?
Masja [62]

Tips for Success

Like any subject, succeeding in mathematics takes practice and patience. Some people find numbers easier to work with than others do. Here are a few tips for working with positive and negative integers:

Context can help you make sense of unfamiliar concepts. Try and think of a practical application like keeping score when you're practicing.

Using a number line showing both sides of zero is very helpful to help develop the understanding of working with positive and negative numbers/integers.

It's easier to keep track of the negative numbers if you enclose them in brackets.

Addition

Whether you're adding positives or negatives, this is the simplest calculation you can do with integers. In both cases, you're simply calculating the sum of the numbers. For example, if you're adding two positive integers, it looks like this:

5 + 4 = 9

If you're calculating the sum of two negative integers, it looks like this:

(–7) + (–2) = -9

To get the sum of a negative and a positive number, use the sign of the larger number and subtract. For example:

(–7) + 4 = –3

6 + (–9) = –3

(–3) + 7 = 4

5 + (–3) = 2

The sign will be that of the larger number. Remember that adding a negative number is the same as subtracting a positive one.

Subtraction

The rules for subtraction are similar to those for addition. If you've got two positive integers, you subtract the smaller number from the larger one. The result will always be a positive integer:

5 – 3 = 2

Likewise, if you were to subtract a positive integer from a negative one, the calculation becomes a matter of addition (with the addition of a negative value):

(–5) – 3 = –5 + (–3) = –8

If you're subtracting negatives from positives, the two negatives cancel out and it becomes addition:

5 – (–3) = 5 + 3 = 8

If you're subtracting a negative from another negative integer, use the sign of the larger number and subtract:

(–5) – (–3) = (–5) + 3 = –2

(–3) – (–5) = (–3) + 5 = 2

If you get confused, it often helps to write a positive number in an equation first and then the negative number. This can make it easier to see whether a sign change occurs.

Multiplication

Multiplying integers is fairly simple if you remember the following rule: If both integers are either positive or negative, the total will always be a positive number. For example:

3 x 2 = 6

(–2) x (–8) = 16

However, if you are multiplying a positive integer and a negative one, the result will always be a negative number:

(–3) x 4 = –12

3 x (–4) = –12

If you're multiplying a larger series of positive and negative numbers, you can add up how many are positive and how many are negative. The final sign will be the one in excess.

Division

As with multiplication, the rules for dividing integers follow the same positive/negative guide. Dividing two negatives or two positives yields a positive number:

12 / 3 = 4

(–12) / (–3) = 4

Dividing one negative integer and one positive integer results in a negative number:

(–12) / 3 = –4

12 / (–3) = –4

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<em><u>In analytic geometry, using the common convention that the horizontal axis represents a variable x and the vertical axis represents a variable y, a y-intercept or vertical intercept is a point where the graph of a function or relation intersects the y-axis of the coordinate system.[1] As such, these points satisfy x = 0.</u></em>

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Here's the solution,

<u>figure</u> 1.

by using trigonometry,

=》

\sin(60)  =   \dfrac{5 \sqrt{3} }{y}

=》

\dfrac{ \sqrt{3} }{2}  =  \dfrac{5 \sqrt{3} }{y}

=》

\dfrac{1}{y}  =  \dfrac{ \sqrt{3} }{2 \times 5 \sqrt{3} }

=》

\dfrac{1}{y}  =  \dfrac{1}{10}

=》

y = 10

And,

=》

\cos(60)  =  \dfrac{w}{y}

=》

\dfrac{1}{2}  =  \dfrac{w}{10}

=》

w = 5

So, values are :

  • y = 10
  • w = 5

<u>figure</u> 2.

Since, it's an isosceles triangle, w = y,

So, by pythagoras theorem :

=》

{w}^{2}  +  {y}^{2}  = (7 \sqrt{2} ) {}^{2}

=》

{w}^{2}  +  {w}^{2}  = 98

=》

2 {w}^{2}  = 98

=》

{w}^{2}  = 49

=》

w =  \sqrt{49}

=》

w = 7

and we know, w = y, so their values will be equal to 7 units.

8 0
3 years ago
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