B= Exactly one solution
-6y+9y= 3y
3y+13=8y-3
3y=8y-16
-5y=-16
First, find the different of the two weights
d = 1,200 - 550
d = 650
The different of the weights is 650
Second, to find the percentage, we should compare the difference to the former weight. So we compare 650 to 1,200, then multiply it to 100%
percentage = 650/1,200 × 100%
percentage = (65,000/1,200) %
percentage = 51.17%
percentage = 51%
The percent of increase is 51%
The 3 numbers which should be placed between 12 and 60 are; 24, 36 and 48.
<h3>What three numbers should be placed between 12 and 60?</h3>
It follows from the task content that the first and last numbers are; 12 and 60 respectively.
Hence, since the difference between 12 and 60 is 48 and there are 4 transitions between the two numbers to have 5 total numbers,
It follows that the common difference is; 48/4 = 12 and the three numbers required are; 24,36 and 48.
Read more on common difference;
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2x2-5x-18=0
Two solutions were found :
x = -2
x = 9/2 = 4.500
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(2x2 - 5x) - 18 = 0
Step 2 :
Trying to factor by splitting the middle term
2.1 Factoring 2x2-5x-18
The first term is, 2x2 its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -18
Step-1 : Multiply the coefficient of the first term by the constant 2 • -18 = -36
Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .
-36 + 1 = -35
-18 + 2 = -16
-12 + 3 = -9
-9 + 4 = -5 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4
2x2 - 9x + 4x - 18
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (2x-9)
Add up the last 2 terms, pulling out common factors :
2 • (2x-9)
Step-5 : Add up the four terms of step 4 :
(x+2) • (2x-9)
Which is the desired factorization
Equation at the end of step 2 :
(2x - 9) • (x + 2) = 0
Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
