8 ones = 1 X 8 = 8
0 tenths = 0 X 1/10 = 0
9 hundredths = 9 X 1/100 = 0.09.
Now we can add everything together to obtain the value.
8 + 0 + 0.09 = 8.09.
Now we can find the difference between the 2 values.
8.09 - 3.40 = 4.69.
4.69 is the final answer.
This pattern of question is always coming up. Since we can't easily guess, then let us set up simultaneous equation for the statements.
let the two numbers be x and y.
Multiply to 44. x*y = 44 ..........(a)
Add up to 12. x + y = 12 .........(b)
From (b)
y = 12 - x .......(c)
Substitute (c) into (a)
x*y = 44
x*(12 - x) = 44
12x - x² = 44
-x² + 12x = 44
-x² + 12x - 44 = 0.
Multiply both sides by -1
-1(-x² + 12x - 44) = -1*0
x² - 12x + 44 = 0.
This does not look factorizable, so let us just use quadratic formula
comparing to ax² + bx + c = 0, x² - 12x + 44 = 0, a = 1, b = -12, c = 44
x = (-b + √(b² - 4ac)) /2a or (-b - √(b² - 4ac)) /2a
x = (-(-12) + √((-12)² - 4*1*44) )/ (2*1)
x = (12 + √(144 - 176) )/ 2
x = (12 + √-32 )/ 2
√-32 = √(-1 *32) = √-1 * √32 = i * √(16 *2) = i*√16 *√2 = i*4*√2 = 4i√2
Where i is a complex number. Note the equation has two values. We shall include the second, that has negative sign before the square root.
x = (12 + √-32 )/ 2 or (12 - √-32 )/ 2
x = (12 + 4i√2 )/ 2 (12 - 4i√2 )/ 2
x = 12/2 + (4i√2)/2 12/2 - (4i√2)/2
x = 6 + 2i√2 or 6 - 2i√2
Recall equation (c):
y = 12 - x, When x = 6 + 2i√2, y = 12 - (6 + 2i√2) = 12 - 6 - 2i√2 = 6 - 2i√2
When x = 6 - 2i√2, y = 12 - (6 - 2i√2) = 12 - 6 + 2i√2 = 6 + 2i√2
x = 6 + 2i√2, y = 6 - 2i√2
x = 6 - 2i√2, y = 6 + 2i√2
Therefore the two numbers that multiply to 44 and add up to 12 are:
6 + 2i√2 and 6 - 2i√2
7. E
8.C
9. F
10.D
11.B
12.A
13.G
I think this is right
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
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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