Answer:
Both A and B are true identities
Step-by-step explanation:
A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n
We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)
So,
n ( n − 2 ) ( n + 2 ) = n(n² - 2²) (difference of two squares)
= n³ - 2²n (expanding the brackets)
= n³ - 4n (simplifying)
So, L.H.S = R.H.S
B. ( x + 1 )² − 2x + y² = x² + y² + 1
We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)
So,
( x + 1 )² − 2x + y² = x² + 2x + 1 - 2x + y² (expanding the brackets)
= x² + 2x - 2x + 1 + y² (collecting like terms)
= x² + 1 + y²
= x² + y² + 1 (re-arranging)
So, L.H.S = R.H.S
So, both A and B are true identities since we have been able to show that L.H.S = R.H.S in both situations.
Answer:
Step-by-step explanation:
What he did at the end of the given equations is solve for x in x + 8y= 21
x = 21 - 8y Substitute that result in the top equation.
<u><em>7(21 - 8y) + 5y = 14 </em></u> is the correct step To continue Remove the brackets
147 - 56y + 5y = 14 Combine
147 - 51y = 14 Add 51y to both sides.
147 = 51y + 14 Subtract 14 from both sides.
133 = 51y divide by 51
y = 2.61 rounded.
The incorrect step is <em><u>underlined and italicized.</u></em>
Notice the grid, the points are (-5, -2) and (4, -1), thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 4}} &,&{{ -1}}~) % (c,d) &&(~{{ -5}} &,&{{ -2}}~) \end{array}~~ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[-5-4]^2+[-2-(-1)]^2}\implies d=\sqrt{(-5-4)^2+(-2+1)^2} \\\\\\ d=\sqrt{(-9)^2+(-1)^2}\implies d=\sqrt{81+1}\implies d=\sqrt{82}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%20-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%7B%7B%20-5%7D%7D%20%26%2C%26%7B%7B%20-2%7D%7D~%29%0A%5Cend%7Barray%7D~~%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-5-4%5D%5E2%2B%5B-2-%28-1%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5-4%29%5E2%2B%28-2%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-9%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B81%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B82%7D)
Answer:
Explanation:
Given the irrational numbers:
In order to arrange the numbers from the least to the greatest, we convert each number into its decimal equivalent.
Finally, sort these numbers in ascending order..
The given numbers in ascending order is:
Note: In your solution, you can make the conversion of each irrational begin on a new line.
8n-28
4(2n-7)
you cant factor this. There are no powers
