Answer:
-9m - 4 + 2m = -9 - 7m + 5
-7m - 4 = -7m - 4
-7m + 7m = -4 + 4
0 = 0
This is an identity solution because the same number equals the same number. That means there's an infinite set of solutions.
Let's solve your equation step-by-step.
Question 1: −2(6−2x) =4(−3+x)
Step 1: Simplify both sides of the equation.
−2(6−2x) =4(−3+x)
(−2) (6) +(−2) (−2x) =(4)(−3)+(4)(x)(Distribute)
−12+4x=−12+4x
4x−12=4x−12
Step 2: Subtract 4x from both sides.
4x−12−4x=4x−12−4x
−12=−12
Step 3: Add 12 to both sides.
−12+12=−12+12
0=0
Answer: All real numbers are solutions.
Question 2:
Let's
solve your equation step-by-step.
5−1(2x+3)
=−2(4+x)
Step 1:
Simplify both sides of the equation.
5−1(2x+3)
=−2(4+x)
5+(−1)
(2x) +(−1) (3) =(−2) (4)+(−2)(x)(Distribute)
5+−2x+−3=−8+−2x
(−2x)
+(5+−3) =−2x−8(Combine Like Terms)
−2x+2=−2x−8
−2x+2=−2x−8
Step 2:
Add 2x to both sides.
−2x+2+2x=−2x−8+2x
2=−8
Step 3:
Subtract 2 from both sides.
2−2=−8−2
0=−10
Answer: There are no solutions.
Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
(3 1/2+8 2/3)-3 5/6
= 1.898717
(10 1/4 - 5 5/8)-1 3/8
= -5.221721
(2 2/3+ 4 5/6)+3 3/8
= 9.795734
If you will simplify this expression you will get this (9x + 3)(6y + 5)
and the factor, which is <span>b)6y+5. </span>
