<span>(4 · 2^5) ÷ (2^3 · 1/16 )
</span>= (2^2 · 2^5) ÷ (2^3 · 2^-4 )
= (2^7) ÷ (2^-1)
= 2^8
ANSWER:
x = 10 / 3
y = 0
STEP-BY-STEP EXPLANATION:
We will be using simultaneous equations to solve this problem. Let's first establish the two equations which we will be using.
Equation No. 1 -
- 6x - 14y = - 20
Equation No. 2 -
- 3x - 7y = - 10
First, we will make ( x ) the subject in the first equation and simplify accordingly.
Equation No. 1 -
- 6x - 14y = - 20
- 6x = - 20 + 14y
x = ( - 20 + 14y ) / - 6
x = ( - 10 + 7y ) / - 3
From this, we will make ( y ) the subject in the second equation and substitute the value of ( x ) from the first equation into the second equation to solve for ( y ) accordingly.
Equation No. 2 -
- 3x - 7y = - 10
- 7y = - 10 + 3x
- 7y = - 10 + 3 [ ( - 10 + 7y ) / - 3 ]
- 7y = - 10 + [ ( - 30 + 21y ) / - 3 ]
- 7y = - 10 + ( 10 - 7y )
- 7y = - 7y
- 7y + 7y = 0
0y = 0
y = 0
Using this, we will substitute the value of ( y ) from the second equation into the first equation to solve for ( x ) accordingly.
x = ( - 10 + 7y ) / - 3
x = [ - 10 + 7 ( 0 ) ] / - 3
x = [ - 10 + 0 ] / - 3
x = - 10 / - 3
x = 10 / 3
The answer would be 1/4 pounds((One and one quarter pounds)) I hope this helps :3
Answer:
x = 1/15
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let's examine the following general product of two binomials with variables x and y in different terms:
so we want the following to happen:
Notice as well that 
means that those two products must differ in just one unit so, one of them has to be negative, or three of them negative. Given that the product 
, then we can consider the case in which one of this (b or d) is the negative factor. So let's then assume that 
are positive.
We can then try combinations for such as:
Just by selecting the first one 
we get that 
and since
This quadratic equation give as one of its solutions the integer: d = -2, and consequently,
Now we have a good combination of parameters to render the factoring form of the original trinomial:
which makes our factorization:
