05/03/2022
05/03/2022 | Evaluate yourself 😊
✅Got it? Good for you.👊🏻
❌Didn't get it? Don't worry, read the explanation below and you'll do better next time😎
Grade 10
1) The first thig we do when we have a fraction raised to a negative exponent, we invert the fraction then make the exponent positive. In other words, we swap the numerator and the denominator and make the exponent positive.
2) When we have a fraction a/b as an exponent, it means we take the b-th root and raise to the power of a. That is why we took the 4-th root and raised to the power of 3.
3) Simplify inside the brackets by getting rid of the 4-th roots and raise the answer to the power of 3.
Grade 11:
1) First thing we do when we solve quadratic inequalities is to simplify our equation to a standard form, that is: ax² + bx + c, and make sure that we only have 0 on our right hand side.
Thus, 3x² + x — 4 ≤ 0 is in standard form.
2) Factorise
(3x + 4)(x — 1) ≤ 0
Just like when you solve for x, without the inequality sign, i.e (3x + 4)(x - 1) = 0, our critical values will be our roots or our x-intercepts of the graph y = 3x² + x — 4. Hence:
x = —4/3 or x = 1
3) Draw your parabola or number line.
Since we are only looking for the x-values where y ≤ 0, it means we want the x-values where the graph is below the x-axis (y = 0 line). Hence, those x-values are between —4/3 and 1.
4) Thus, our solution:
—4/3 ≤ x ≤ 1
Grade 12
1) A geometric convergent series is a series that has a limit or approaches a certain constant. This happens only when the constant ratio of the series is between —1 and 1, that is —1 < r < 1
2) Find your constant ratio, substitute in the inequality
—1 < r < 1 then solve for x
3) After solving for x:
—1 < 2x + 3 < 1, we find that the series will converge when x is between —2 and 1.
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