We'll use the exponent rule that a^(b+c) = a^b*a^c
to break up 17^16 into 17^(14+2) = 17^14*17^2
That allows us to factor like so
17^16+17^14
17^14*17^2+17^14*1
17^14(17^2 + 1)
Computing 17^14 is a lot more difficult compared to computing 17^2 + 1, assuming you only could use paper and pencil.
Note that 17^2 = 17*17 = 289, so 17^2 + 1 = 289+1 = 290
Because 290 ends in a 0, this means that 5 is a factor
290 = 5*58
This makes 5 to be a factor of the original expression.
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So overall,
17^16+17^14 = 17^14(17^2 + 1) = 17^14*290 = 17^14*5*58 = 5*17^14*58
The stuff "17^14*58" doesn't matter. All we care about is that factor of 5.
Since 5 is a factor, this means that 17^16+17^14 is a multiple of 5, or 17^16+17^14 is divisible by 5.
Phrased another way, 17^16+17^14 all divided over 5 leads to some integer. In this case, the integer would be the result of 17^14*58.
Answer:
-9/50
Step-by-step explanation:
Answer:
5cm wide, 10cm long.
Step-by-step explanation:
I came to this answer by looking for common factors between 50 and 5, and I found 10.
5cm + 5cm = 10cm, so it works.
To confirm this, I multiplied 5 and 10 and got 50cm (squared) which works.
Answer: 0.75
Step-by-step explanation: took the same question
Answer:
The solution for the system of linear equations 3x-y=10 and 2x+y=5 is x=3 y= -1
<u>Solution:
</u>
Given that two linear equations are 3x-y = 10 and 2x + y = 5
We have to find the values of “x” and “y”
Let us consider 3x – y =10 ---- eqn 1
2x + y = 5 --- eqn 2
From eqn 1 , rearranging the terms we get
y = 3x-10 --- eqn 3
By substituting the value of “y” from eqn 3 into eqn 2 we get,
2x + 3x – 10 = 5
On solving above expression, 5x – 10 = 5
5x = 15
x = 3
Substitute the value of “x” in eqn 1 to obtain “y” value
3(3) – y = 10
9 – y = 10
y = 9-10 = -1
Hence the solution for the system of linear equations 3x-y=10
and 2x+y=5 is x=3 and y= -1
