Beforehand let me apologize for my sloppy handwriting. I'm a lefty, so please deal with me. . Anyway, the equation you listed didn't really have a x, because a linear equation is y= mx+b, and anything with x is the slope. But I did see y and a number on one side, so I'm like maybe I can get these two alone. So here's what I did:
2y + 4=0
-4 -4 I subtracted 4 on both sides. Why because you would want "y" alone.
Next,
2y= -4
Now to get "y" alone you want to divide on both sides.
2y= -4/ 2
y= -2
Now you're probably thinking "how do you graph it? There's no "x" in the equation." Well, you just graph it. Since the answer is y= -2, you go to the y axis, look for -2, and place a line to indicate that is the equation. And to make it clear, remember y is the y-intercept.
I really do hope this helps you, if not message me. I'll be happy to help, and again I'm sorry for my handwriting!
You order them least to greatest, then take out your mean, median, and mode. Then you've got your outliners left over.
Answer:
A) 0
Step-by-step explanation:
When x is divided by 11, we have a quotient of y and a remainder of 3
x/11 = y + 3
x = 11y + 3 ........(1)
When x is divided by 19, we have a remainder of 3 also
x/19 = p + 3 (p = quotient)
x = 19p + 3 ..........(2)
Equate (1) and (2)
x = 11y + 3 = 19p + 3
11y + 3 = 19p + 3
11y = 19p + 3 -3
11y = 19p
Divide both sides by 11
11y/11 = 19p/11
y = 19p/11
y and p are integers. 19 is a prime number. P/11 is also an integer
y = 19(integer)
This implies that y is a multiple of 19. When divided by 19, there is no remainder. The remainder is 0
Answer:
Step-by-step explanation:
14 + 5t > = 49 (thats a greater then or equal sign)
5t > = 49 - 14
5t > = 35
t > = 35/5
t > = 7 <===
Answer with explanation:
→→→Function 1
f(x)= - x²+ 8 x -15
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= - 2 x + 8
Put,f'(x)=0
-2 x+ 8=0
2 x=8
Dividing both sides by , 2, we get
x=4
Double differentiating the function
f"(x)= -2, which is negative.
Showing that function attains maximum at ,x=4.
Now,f(4)=-4²+ 8× 4-15
= -16 +32 -15
= -31 +32
=1
→→→Function 2:
f(x) = −x² + 2 x − 3
Differentiating once , to obtain Maximum or minimum of the function
f'(x)= -2 x +2
Put,f'(x)=0
-2 x +2=0
2 x=2
Dividing both sides by , 2, we get
x=1
Double differentiating the function,gives
f"(x)= -2 ,which is negative.
Showing that function attains maximum at ,x=1.
f(1)= -1²+2 ×1 -3
= -1 +2 -3
= -4 +2
= -2
⇒⇒⇒Function 1 has the larger maximum.