Answer:
#1. A. C. D.
#2. D. F.
3^4 equals 81 and so do answer D. & F.
Answer:
The decimal number system in use today was first recorded in Indian mathematics.
Step-by-step explanation:
In the classical period of Indian mathematics (400 AD to 1200 AD), important contributions were made by scholars like Aryabhata, Brahmagupta, Bhaskara II, and Varāhamihira. The decimal number system in use today was first recorded in Indian mathematics.
<u>2x + 3y = 1</u>
<u>y = 3x + 15</u>
There's not much you can do with the first equation, because it has
two variables in it ... 'x' and 'y' . No matter how much you move them
around, you'll never be able to get either one equal to just a number.
Is there any way you could get rid of one of the variables in the first
equation, and have just 1 letter in it to solve for ?
Absolutely ! The second equation tells you something that 'y' is <u>equal</u> to,
(3x + 15). "EQUAL" is very powerful. It means that wherever you see 'y',
you can put (3x + 15) in its place, and you won't change anything or
upset anything. One thing you can do is take that (3x + 15) from the <span>
2nd</span> equation, and put it right into the first equation in place of 'y'.
You'll see how that helps as soon as you do it.
First equation: <u>2x + 3y = 1</u>
Substitute for 'y' : 2x + 3(<em>3x + 15</em>) = 1
Remove parentheses: 2x + 3(3x) + 3(15) = 1
2x + 9x + 45 = 1
Combine the terms with 'x' in them: 11x + 45 = 1
Look what you have now ! An equation with only one variable in it !
Subtract 45 from each side: 11x = -44
Divide each side by 11 : <em> x = -4</em>
You're more than halfway there. Now you know what 'x' is,
and you can use it with either equation to find what 'y' is.
-- If you use it with the first equation: <u> 2x + 3y = 1</u>
Put in the value of 'x': 2(<em>-4</em>) + 3y = 1
Remove the parentheses: -8 + 3y = 1
Add 8 to each side: 3y = 9
Divide each side by 3 : <em> y = 3</em>
-- If you use it with the 2nd equation: <u>y = 3x + 15</u>
Put in the value of 'x' : y = 3(<em>-4</em>) + 15
Remove the parentheses: y = -12 + 15
Add numbers on the right side: <em> y = 3</em> (same as the other way)
So there's your solution for the system of two equations:
<em> x = -4</em>
<em> y = 3</em>
Given:

To find:
The set of points that includes all of the solutions for the given equation.
Solution:
From the given equation, it is clear that y-values depend on the x-values.
If x is a real number, then y-values are real numbers such that
.
So, the set of points that includes all of the solutions for the given equation is defined as
for all real numbers.
Therefore, the correct option is B.
<em><u>5</u></em><em><u>0</u></em><em><u> </u></em><em><u>per</u></em><em><u> </u></em><em><u>cent</u></em><em><u> </u></em><em><u>can</u></em><em><u> </u></em><em><u>also</u></em><em><u> </u></em><em><u>be</u></em><em><u> </u></em><em><u>termed</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>half</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>number</u></em><em><u>.</u></em>
<em><u>therefore</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>½</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>original</u></em><em><u> </u></em><em><u>number</u></em>
<em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>multiplied</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>2</u></em>
<em><u>answer</u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u>0</u></em>