<span>81x2</span> as <span><span>(9x)</span>2</span>.<span><span><span>(9x)</span>2</span>−49
</span>Make 49 as 72.<span><span><span>(9x)</span>2</span><span>−72</span></span>
<span><span>a2<span>−b2</span></span>=<span><span>(a+b)</span><span>(<span>a−b</span>)</span></span></span>
when <span>a=9x</span> and b=7.<span><span>(<span>9x+7</span>)</span><span>(<span>9x−7</span>)</span></span>
The problem would be
2/3 divided by 5/3
That’s because 2/3 times 3/5 = 2/3 divided by 5/3
(When dividing by a fraction you multiply by its reciprocal)
Hope this helps :)
The values In wxyz are mathematically given as
W=4
y=5
X=0
<h3>What is the values In
XYZ?</h3>
- My one's digit and my tens digit together make up a total of seven.
- My one digit has three more digits than my tens digit does.
Let the equation for the number be: wxyz.
- Digits at tens place = yz
- Diglt at ones place = z
Generally, the equation for is mathematically given as
z+y=7 ......1
y=2+3 .......2
solving further
z+3+z-7
2z 7-3
z=2
Putting the value of z Within the second equation:
y=z+3 y=2+3
y=5
1. The digit in the thousands place of my number is 4, and
2. The digit in the hundreds place is four fewer than the digit in the thousands place.
W=4
X=w-4
x=4-4
X=0
In conclusion, the values In wxyz are
W=4
y=5
X=0
Read more about digits
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A. The student improperly set up the equation to find the y-intercept
This is because he put y as -2 and x as 1, but the coordinate is (1,-2) not (-2,1)
Hope this helps!