Two fractions are equivalent if the product of the numerator of the first and the denominator of the second is equal to the product of the numerator of the second and the denominator of the first.
4 x 2 ___ 16 x 0.25
8 is not equal to 4
Thus, they are not equivalent.
Answer:
no solution
Step-by-step explanation:
For getting the nature of solution of the quadratic equation of the form:
ax² + bx + c = 0
We need to find Discriminant which is:
Discriminant (D) = b² - 4ac
- If D < 0, there is no solution of equation.
- If D = 0, there are two equal and real solution of equation
- If D < 0, there are two real and distinct solution of equation
Here we have equation is:
2x² - 9x + 12 = 0
∴ a=2, b = -9, c = 12
⇒ D = 81 - 4 × 2 × 12 = -16 < 0
Hence, there is no solution of given equation.
If B is the midpoint of AC, then |AB| = |AC|.
|AB| = 3x + 2
|BC| = 5x - 10
Therefore we have the equation:
3x + 2 = 5x - 10 |subtract 2 from both sides
3x = 5x - 12 |subtract 5x from both sides
-2x = -12 |divide both sides by (-2)
x = 6
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Answer:
The equation is 
Step-by-step explanation:
Linear function:
A linear function for the amount of money in an account after t days is given by:
In which A(0) is the initial value and m is the daily cost.
Your EZ Pass account begins with $80. It costs you $4/day.
This means that 
So
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
