Well, let's first solve each equation:
1.) -4x + 6 - 3x = 12 - 2x - 3x
To start, combine each like-term on each side of the equal sign (The numbers with variables in-common // the numbers alike on the same side of the equal sign):
-7x + 6 = 12 - 5x
Now, we get the two terms with variables attached to them, on the same side, so, we do the opposite of subtraction, which is, addition:
-7x + 6 = 12 - 5x
+5x +5x
_____________
-2x + 6 = 12
Next, you do the opposite of addition, which is, subtraction, and, subtract 6 from both sides:
-2x + 6 = 12
-6 -6
____________
-2x = 6
Finally, divide by -2 on each side, to find out what the value of 'x' is:
-2x = 6
÷-2 ÷-2
________
x = -3
So, the answer is not 'A.'
_________________________________________
Now, we test out the rest of the equations, the exact same way:
2.) 4x + 6 + 3x = 12 + 2x + 3x
Combine your like-terms, on each side of the equal sign:
7x + 6 = 12 + 5x
Now, get both terms, with the variable, 'x,' to the same side, and, to do that, do the opposite of addition, which is, subtraction:
7x + 6 = 12 + 5x
-5x -5x
______________
2x + 6 = 12
Next, subtract 6 from both sides:
2x + 6 = 12
-6 -6
__________
2x = 6
Finally, divide by 2, on both sides:
2x = 6
÷2 ÷2
__________
x = 3
So, the answer is 'B.'
_________________________________________
3.) 4x + 6 - 3x = 12 - 2x - 3x
Again, we combine the like-terms, on both sides of the equal sign:
x + 6 = 12 - 5x
Now, we get both terms with the variable 'x,' to the same side, and, the opposite of subtraction, is addition, so, we're going to add 5x to both sides:
x + 6 = 12 - 5x
+ 5x + 5x
______________
6x + 6 = 12
Now, we subtract 6 from each side, because, the opposite of addition, is subtraction:
6x + 6 = 12
- 6 - 6
_____________
6x = 6
Now, divide by 6, on both sides:
6x = 6
÷ 6 ÷ 6
_____________
x = 1
So, the answer is not 'C.'
_________________________________________
4.) 4x + 6 - 3x = 12x + 2x + 3x
First, we combine the like-terms:
x + 6 = 17x
Next, we get both terms, with the variable, 'x,' to the same side:
x + 6 = 17x
-x -x
_____________
6 = 16x
Now, divide by 16, on both sides:
X = 3/8
So, 'D,' is not the answer.
_______________________
The answer is, 'B.'
I hope this helps!
The <span>perimeter of a face of the new cube is 8 times larger than the previous cube.
</span>Let us say, the side of cube before increase is "a"
Then perimeter of the cube = 12a ,
because the cube has 12 sides.
And perimeter of a face of cube = 4a
Now, the side length is increased by factor of 3 i.e. new side is 3a
Thus, perimeter of new face of the cube = 4*3a = 12a
Therefore, the new perimeter is larger by = 12a - 4a = 8a
Explanation:
The question requires that we round the given number to Ten-thousand
To do so, we will have to write the place values of the numbers first
Next, we find the digits before the Ten-thousand digits and then approximate the value
One of the principles to use is that
numbers from 0 to 4 are rounded to 0
while numbers from 5 to 9 are rounded to 1
In our case, The thousand digit is 0, so we round down to 0, then we add to the Ten-thousand digits and convert each of the numbers before the Ten-thousand digits to 0
Thus, we will have
Hence, the answer is 1,370,000
At the start, the tank contains A(0) = 50 g of salt.
Salt flows in at a rate of
(1 g/L) * (5 L/min) = 5 g/min
and flows out at a rate of
(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 5 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 5
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)
(e^(t/40) A(t))' = 5e^(t/40)
e^(t/40) A(t) = 200e^(t/40) + C
A(t) = 200 + Ce^(-t/40)
Given that A(0) = 50, we find
50 = 200 + C ==> C = -150
so that the amount of salt in the tank at time t is
A(t) = 200 - 150 e^(-t/40)
