The <em>approximate</em> solution of the equation shown in the picture is x ≈ 39 / 8 (Right choice: B).
<h3>How to find an approximate solution of a one-variable equation</h3>
The solution of the equation is between x = 4 and x = 5. Now we begin by evaluating each side of the expression (f(x) = x² - 5 · x + 4, g(x) = 2 / (x - 1)) at the average of x = 4 and x = 5.
x = (4 + 5) / 2
x = 4.5
f(4.5) = 4.5² - 5 · 4.5 + 1
f(4.5) = - 5 / 4
g(4.5) = 2 / (4.5 - 1)
g(4.5) = 4 / 7
The solution of the equation is between x = 4.5 and x = 5, then we evaluate at the average:
x = (4.5 + 5) / 2
x = 4.75
f(4.75) = 4.75² - 5 · 4.75 + 1
f(4.75) = - 3 / 16
g(4.75) = 2 / (4.75 - 1)
g(4.75) = 8 / 15
The solution of the equation is between x = 4.75 and x = 5, then we evaluate at the average:
x = (4.75 + 5) / 2
x = 4.875
f(4.875) = 4.875² - 5 · 4.875 + 1
f(4.875) = 25 / 64
g(4.875) = 2 / (4.875 - 1)
g(4.875) = 16 / 31
The <em>approximate</em> solution of the equation shown in the picture is x ≈ 39 / 8 (Right choice: B).
To learn more on successive approximations: brainly.com/question/27191494
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Use this formula: C=2πr. Divide 14/2 to get the radius. Substitute the radius into the formula.
The answer is 14pi units.
<u>Please mark this answer as the Brainliest if it helped! Thank you.</u>
85700, 8570, 857, 85.7, 8.57, .857
Because you have to go in the order of PEMDAS, parenthesis, exponents, multiplication, division, addition, and subtraction. From what you are saying, 'why you must simplify 3 to the second power first' it seems that you have no parenthesis, so you move onto what is next, which is exponents, in this case, 3 to the power of two.
(<u>−1</u>
2 )(n^3)+
<u>1</u>
2 n^2+4.6n+(−
<u>1</u>
2)(n^3)+
<u>1</u>
2 n^2+4.5n
=
<u>−1</u>
2 n^3+
1
2 n^2+4.6n+
−1
2 n^3+
1
2 n^2+4.5n
Combine Like Terms:
=
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.6n+
<u>−1</u>
2 n^3+
<u>1</u>
2 n^2+4.5n
=(<u>−1</u>
2 n^3+
<u>−1</u>
2 n^3)+(
<u>1</u>
2 n^2+
<u>1</u>
2 n^2)+(4.6n+4.5n)
=−n^3+n^2+9.1n
Answer:
=−n^3+n^2+9.1n
Everything underlined means its a fraction/divided hope this helps <em>:D</em>
