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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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:
D
Step-by-step explanation:
I took the test
1a) f(x) = I x+2 I. This is a piece-wise graph ( V form)
x = 0 →f(x) =2 (intercept y-axis)
x = -2→f(x) = 0 (intercept x-axis)
x = -3→f(x) = 1 (don't forget this is in absolute numbers)
x = -4→f(x) = 2 (don't forget this is in absolute numbers)
Now you can graph the V graph
1b) Translation: x to shift (-3) units and y remains the same, then
f(x-3) = I x - 3 + 2 I = I x-1 I
the V graph will shift one unit to the right, keeping the same y. Proof:
f(x) = I x-1 I . Intercept x-axis when I x-1 I = 0, so x= 1
