There are two numbers whose sum is 64. The larger number subtracted from 4 times the smaller number gives 31. Then the numbers are 45 and 19
<h3><u>
Solution:</u></h3>
Given that, There are two numbers whose sum is 64.
Let the number be a and b in which a is bigger.
Then, a + b = 64 ------ eqn (1)
The larger number subtracted from 4 times the smaller number gives 31.
4 x b – a = 31
4b – a = 31 ----- eqn (2)
We have to find the numbers.
So, from eqn (2)
a = 4b – 31
Subatitute a in (1)
4b – 31 + b = 64
On solving we get
5b = 64 + 31
5b = 95
b = 19
So, b = 19, then eqn 1
a + 19 = 64
On simplification,
a = 64 – 19
a = 45
Hence, the two numbers are 45 and 19
Answer:
28980
Step-by-step explanation:
Also "<em>nice</em>"
The square root of a a negative integer is imaginary.
It would still be a negative under a square root if you multiplied it by 2, therefor it will still be imaginary, or I’m assuming as your book calls it, undefined.
2•(sqrt-1) = 2sqrt-1
If you add a number to -1 itself, specifically 1 or greater it will become a positive number or 0 assuming you just add 1. In that case it would be defined.
-1 + 1 = 0
-1 + 2 = 1
If you add a number to the entire thing “sqrt-1” it will not be defined.
(sqrt-1) + 1 = 1+ (sqrt-1)
If you subtract a number it will still have a negative under a square root, meaning it would be undefined.
(sqrt-1) + 1 = 1 + (sqrt-1)
however if you subtract a negative number from -1 itself, you end up getting a positive number or zero. (Subtracting a negative number is adding because the negative signs cancel out).
-1 - -1 = 0
-1 - -2 = 1
If you squared it you would get -1, which is defined
sqrt-1 • sqrt-1 = -1
and if you cubed it, you would get a negative under a square root again, therefor it would be undefined.
sqrt-1 • sqrt-1 • sqrt-1 = -1 • sqrt-1 = -1(sqrt-1)
Sorry if this answer is confusing, I don’t have a scientific keyboard, I’ll get one soon.
Answer:
Jacob served 16 quarts of juice.
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
radius squared * pi = 63.585
