Terms & Expressions Overview
A brief quiz on all parts of the course book "Terms and Expressions".
A brief quiz on all parts of the course book "Terms and Expressions".
Kartei Details
| Karten | 12 |
|---|---|
| Lernende | 20 |
| Sprache | English |
| Kategorie | Mathematik |
| Stufe | Mittelschule |
| Erstellt / Aktualisiert | 06.11.2023 / 09.11.2023 |
| Lizenzierung | Keine Angabe |
| Weblink |
https://card2brain.ch/box/20231106_terms_expressions_overview
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| Einbinden |
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If we evaluate the expression \(T(x) = x^3+1\) for \(x=2 \), what do we get?
\(T(2) =\)
9
Wieso dürfen wir beim Doppelbruch \(\frac{\frac{p}{3}}{\frac{3}{4}}\)nicht folgendes tun?
\(\frac{\frac{p}{3}\cdot3}{\frac{3}{4}\cdot4}\)
Zwar würden so die Nenner in beiden Brüchen wegfallen, aber wir haben den Wert des Doppelbruchs verändert! Das dürfen wir natürlich nicht, wir müssen immer oben und unten mit der selben Zahl multiplizieren.
With which factors do we have to expand each fraction to obtain a common denominator?
\(\frac{1}{x+1}+\frac{2}{y}+\frac{1}{x}\)
The LCM of the denominators is \(xy(x+1)\). Therefore, we expand the first fraction with \(\frac{xy}{xy}\), the second one with \(\frac{x(x+1)}{x(x+1)}\), and the third with \(\frac{y(x+1)}{y(x+1)}\).
Is the following calculation correct?
\(\frac{a}{x}+\frac{b}{y} = \frac{ay+bx}{xy}\)
Yes!
We have to expand the fractions to the LCM of the denominators. Here, \(LCM(x,y)=xy\) .
Then, \(\frac{a}{x}+\frac{b}{y} =\frac{a}{x}\cdot\frac{y}{y}+\frac{b}{y}\cdot\frac{x}{x}= \frac{ay}{xy}+\frac{bx}{xy} = \frac{ay+bx}{xy}\).
Is the following calculation correct?
\(\frac{3x-1}{3y} = \frac{x-1}{y}\)
No!
-1 was discriminated and was not divided by 3, as it should have been.
Is this calculation correct?
\(\frac{1-3x}{3x-1} = -1\)
Yes, it is correct:
\(\frac{1-3x}{3x-1} = \frac{(1-3x)}{(-1)(1-3x)} = \frac{1}{-1} =-1\)
What is the \(LCM(2^3\cdot3^3, 2^2\cdot6^2)\)?
Careful, 6 is not a prime number! Therefore, we have to find its prime factors first: \(6^2 =(2\cdot3)^2 = 2^2\cdot3^2\).
Thus, the LCM becomes:
\(LCM(2^3\cdot3^3, 2^2\cdot6^2)=LCM(2^3\cdot3^3, 2^4\cdot3^2) = 2^4\cdot3^3\)
What is the \(LCM(a^2-b^2, a+b, a)\)?
First, we factor out all expressions. The brackets around (a+b) are not needed, but they help to see that the whole expression is one factor.
Then, we account for all available factors:
\(LCM(a^2-b^2, a+b, a) = LCM((a-b)(a+b),(a+b), a) =a(a-b)(a+b)\)
