What?? I’m sorry what are you asking?
Answer:m=2
Step-by-step explanation:
A. Alright, we want to multiply one equation by a constant to make it cancel out with the second. Since the first equation has a "blank" y, let's multiply the first equation by <em>2</em>.
3x-y=0 → 2(3x-y=0) = 6x - 2y = 0
5x+2y=22
The answer for this part would be: 6x - 2y = 0 and 5x + 2y = 22
B. So now we combine them:
6x - 2y = 0
+ + +
5x + 2y = 22
= = =
11x + 0 = 22 ← The answer
C. Now that we have the equation 11x = 22, we solve for x
11x = 22 ← Divide both sides by 11
x = 2 ← The answer
D. Now that we have x=2, we plug that back in to 5x+2y=22 and solve for y:
5(2)+2y = 22
10 + 2y = 22
2y = 12
y = 6
<u>Therefore, the solution to this problem is x = 2 and y = 6</u>
Answer:
a. ∀ x≤0 ∧ ∀ y ∈ IR + ∪ 0
b. ∀ x≤0 ∧ ∀ y ∈ IR -
c. ∀ x≤0 ∧ ∀ y -∞ → +∞
Step-by-step explanation:
P=4l ⇒ 28=4l ⇔ l=28/4 =7
A=l.l=7*7=49
As can be seen in the graph so that x is negative, it must be fulfilled that its values are located in the second and third quadrants and their values must be less than zero and also the values of y can be all positive and negative real numbers
Possible coordinates:
a. ∀ x≤0 ∧ ∀ y ∈ IR + ∪ 0
for all x less iqual than zero and all positive real y including zero
b. ∀ x≤0 ∧ ∀ y ∈ IR -
for all x less iqual than zero and all y negative reals
c. ∀ x≤0 ∧ ∀ y -∞ → +∞
for all x less equal than zero and all y from minus infinite to plus infinite