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:
98
Step-by-step explanation:
56 x 1 3/4
56 x 1.75 = 98
Answer:
-3.2
Step-by-step explanation:
Do you have to work this out on paper because I just used a calculator but just add 3.1 + -6.3.
Answer:
1. (3x - 2)(x + 5).
2. 2(x + 4)(x - 2).
3. (2x - 9)(2x + 9).
Step-by-step explanation:
1. f(x) = 3x^2 +13x -10
Try to factor by the ac method:
3 * -10 = -30
We need 2 numbers whose product is -30 and whose sum is + 13.
They are -2 and +15, so we write:
3x^2 + 15x - 2x - 10 Factor by grouping:
= 3x(x + 5) - 2(x + 5)
= (3x - 2)(x + 5).
2. f(x) = 2x^2+4x-16
= 2(x^2 + 2x - 8)
Now we need 2 numbers whose product is -8 and whose sum is + 2.
They are +4 and -2. So:
= 2(x + 4)(x - 2).
3. 4x^2 - 81 This is the difference of 2 squares so the factors are:
(2x - 9)(2x + 9).
-72yexponet73 + 2xexponet2 :)
