We have two fractions, \dfrac{1}{6} 6 1 start fraction, 1, divided by, 6, end fraction and \dfrac{3}{8} 8 3 start fraction,
1 answer:
Answer:
<em>B) 24</em>
<em>D) 48</em>
<em></em>
Step-by-step explanation:
Given:
Two fractions
To find:
Number that can be chosen as Common denominator such that numerator is also a whole number ?
Solution:
Common denominator for two fractions and
is chosen as LCM or multiple of LCM of (q, s).
OR
Common denominator for two fractions is chosen as the Least Common Multiple or multiple of LCM of denominators of the two fractions.
The denominators of the given fractions are 6 and 8.
Let us factorize and try to find the LCM of 6 and 8.
Common part of the denominators (as underlined) will be taken only once.
So,
Multiples of LCM, 24 = 48
So, the correct answers are:
<em>B) 24 </em>and
<em>D) 48</em>
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Answer:
We want to find a solution of the system:
x^2 + y^2 > 49
y ≤ –x^2 – 4
Here we do not have any options, so let's try to find a general solution.
First, we can remember that the equation of a circle centered in the point (a, b) and of radius R is:
(x - a)^2 + (y - b)^2 = R^2
If we look at our first inequality, we can write it as:
x^2 + y^2 > 7^2
So the solutions of the first inequality are all the points that are outside (because the symbol used is >) of the circle of radius R = 7 centered in the origin.
From the other equation, we would get:
y ≤ –x^2 – 4
This is parabola, anything that is in the graph of the parabola or below will be a solution for this inequality.
Then the solutions of the system, are the ones that are in the region of solutions for both inequalities.
You can see the graph below, where both regions are graphed. The intersection of these regions is the region of the solutions for the system of inequalities:
by looking at the graph, we can see a lot of points that are solutions, like:
(0, -10)
(0, -15)
(2, -10)
etc.
