Answer:
9
Step-by-step explanation:
6/2(1+2)
6/2(3)
3(3)
9
I hope this helped! If so, please mark brainliest!
Answer:
x(2x+3)
Step-by-step explanation:
Im guessing 2x2 is 2x^2
2x^2 + 3x = 0
x(2x+3)
Since the nickles are first and there are (1.00 /.05= 20) 20 nickles in a dollar the first number is 20 then next is quarters and there is (1.00 / .25= 4) 4 quarters in a dollar so the second number is 4 so now u have to simplify if you were told to
the number 20 can be divided by 4 and the answer is 5 the number 4 is next divide by 4 and is 1 so the whole answer is <u>5 to 1 or 5:1</u> <u /><u />
Answer:100
Step-by-step explanation:im not sure but I think it’s 100
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.
