Answer:
Tom’s age is 7 years
Mary’s age is 13 years
Step-by-step explanation:
Since we do not know the ages, let’s represent the ages by variables at first.
Let m represent mary’s age will t represent Tom’s age.
Now, let’s proceed to have equations.
Adding square of Tom’s age (t^2) to mary’s age give 62
t^2 + m = 62 •••••••(i)
Adding square of mary’s age (m^2) to Tom’s age give 176
m^2 + t = 176 •••••••(ii)
Now, to get the individual ages, we will need to solve both equations simultaneously.
Solving both equations simultaneously without mathematical softwares can be a little hard.
By the use of mathematical software ( wolfram alpha to be specific), we can input both equations and allow the software to solve.
By inputing these equations, we have the values of t to be 7 and m to be 13
And if we try to check by inspection, we can see that these values are actually correct.
7^2 + 13 = 62
13^2 + 7 = 176
Answer:
you too
Step-by-step explanation:
Answer:
See below.
Step-by-step explanation:
First, notice that this is a composition of functions. For instance, let's let 
and 
. Then, the given equation is essentially 
. Thus, we can use the chain rule.
Recall the chain rule: ![\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%3Df%27%28g%28x%29%29%5Ccdot%20g%27%28x%29)
. So, let's find the derivative of each function:
We can use the Power Rule here:
Now:
Again, use the Power Rule and Sum Rule
Now, we can put them together:
Answer:
x>3
Hope this helps!
By way of example, suppose <em>A</em> = {1, 2, 3} and <em>B</em> = {<em>a</em>, <em>b</em>, <em>c</em>}. Then the Cartesian product of <em>A</em> and <em>B</em> is
<em>A</em> × <em>B</em> = {{1, <em>a</em>}, {1, <em>b</em>}, {1, <em>c</em>}, {2, <em>a</em>}, {2, <em>b</em>}, {2, <em>c</em>}, {3, <em>a</em>}, {3, <em>b</em>}, {3, <em>c</em>}}
That is, each element in <em>A</em> gets a pairing with each element in <em>B</em>, and for each pairing you have <em>n(A)</em> choices for the first element and <em>n(B)</em> choices for the second element.
So if <em>n(A)</em> = <em>p</em> and <em>n(B)</em> = <em>q</em>, then <em>n(A</em> × <em>B)</em> = <em>pq</em>.
