Answer:
21 and 3
Step-by-step explanation:
Let's call the small number and the larger number .
According to the problem:
And when you substract the smaller number from the larger number you get 18:
substituting the first equation in the second one:
the larger number is 21.
Using the first equation to find :
the smaller number is 3.
Rounding 0.64 to the nearestwholenumber gives us 1.
We have,
The concept used to round 0.64 to the nearestwholenumber is the concept of rounding.
When rounding a decimal number, we look at the digit immediately to the right of the decimal point.
If that digit is 5 or greater, we round up to the next wholenumber.
If the digit is less than 5, we round down to the current whole number.
In this case, the digit to the right of the decimal point is 4, which is less than 5.
Therefore, we round down to the nearest whole number, which is 0.
To round 0.64 to the nearestwholenumber, we look at the digit immediately to the right of the decimal point, which is 6.
Since 6 is greaterthan 5, we round down.
Therefore,
Rounding 0.64 to the nearestwholenumber gives us 1.
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Answer:
first X=8.94
second X=34
Step-by-step explanation:
5) x^2=4^2+8^2
x^2=16+64=80
x=8.94
6)x^2= 31^2+14^2
x^2=961+196
x^2=1157
x=34
Answer:
3.
Step-by-step explanation:
This is a geometric series so the sum is:
a1 * r^n - 1 / (r - 1)
= 1 * (2^101 -1) / (2-1)
= 2^101 - 1.
Find the remainder when 2^101 is divided by 7:
Note that 101 = 14*7 + 3 so
2^101 = 2^(7*14 + 3) = 2^3 * (2^14)^7 = 8 * (2^14)^7.
By Fermat's Little Theorem (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.
So 2^101 mod 7 = (8 * 4^7) mod 7
= (8 * 4) mod 7
= 32 mod 7
= 4 = the remainder when 2^101 is divided by 7.
So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..