It seems that some the work is already here, but I'd be glad to!! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that's as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
Hope this helps!
Answer: "Count 2 and then count 31 more."
Step-by-step explanation:
We have the equation:
231 - 198
Now, the negative number is kinda ugly, so i will write it as:
200 - 2 = 198
Is a lot easier work with 200 and 2, than with 198.
Then the equation is now:
231 - (200 - 2)
And the left number we also have a "200", so it can be written as:
200 + 31 = 231
As this is a sum, we can ignore the parentheses.
200 + 31 - 200 + 2
31 + 2
Then the correct option is:
"Count 2 and then count 31 more."
Answer:
x=1 or x=−2
Step-by-step explanation:
16x2+10x−27=−6x+5
Step 1: Subtract -6x+5 from both sides.
16x2+10x−27−(−6x+5)=−6x+5−(−6x+5)
16x2+16x−32=0
Step 2: Factor left side of equation.
16(x−1)(x+2)=0
Step 3: Set factors equal to 0.
x−1=0 or x+2=0
x=1 or x=−2
104/-120
=-13/15
Devishri1977 gave a better example of how she got to the answer
X=4 is the answer to this
