The function f(x)=x−5 x 2−25 is defined everywhere except at x=±5. if possible, define f at x=5 so that it becomes continuous at
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The triangle to solve for has a leg or 12 feet, hypotenuse of 20 feet, so solve for Pythagoras 
for a
a = 16 feet
a = 16 feet
Hello!
Since we already see an equation that equals y (-2x-4), we can plug that equation into the first to solve for x.
x+4(-2x-4)=19
x-8x-16=19
-7x=35
x=-5
Now that we know the value of x, we will plug it into the second equation to solve for y.
y=-2(-5)-4
y=10=4
y=6
This gives us the answer below
The correct answer is A) (-5,6)
I hope this helps!
Since we already see an equation that equals y (-2x-4), we can plug that equation into the first to solve for x.
x+4(-2x-4)=19
x-8x-16=19
-7x=35
x=-5
Now that we know the value of x, we will plug it into the second equation to solve for y.
y=-2(-5)-4
y=10=4
y=6
This gives us the answer below
The correct answer is A) (-5,6)
I hope this helps!