The like terms are 12 and 5, so you have to subtract 5 from both sides.
12 < x + 5
-5. -5
7 < x
We can just switch it around so x is in the left side.
x > 7
Centre= (-1,-2)
Radius=5
All working shown on photo
Answer:
-1/8
Step-by-step explanation:
lim x approaches -6 (sqrt( 10-x) -4) / (x+6)
Rationalize
(sqrt( 10-x) -4) (sqrt( 10-x) +4)
------------------- * -------------------
(x+6) (sqrt( 10-x) +4)
We know ( a-b) (a+b) = a^2 -b^2
a= ( sqrt(10-x) b = 4
(10-x) -16
-------------------
(x+6) (sqrt( 10-x) +4)
-6-x
-------------------
(x+6) (sqrt( 10-x) +4)
Factor out -1 from the numerator
-1( x+6)
-------------------
(x+6) (sqrt( 10-x) +4)
Cancel x+6 from the numerator and denominator
-1
-------------------
(sqrt( 10-x) +4)
Now take the limit
lim x approaches -6 -1/ (sqrt( 10-x) +4)
-1/ (sqrt( 10- -6) +4)
-1/ (sqrt(16) +4)
-1 /( 4+4)
-1/8
Answer:
its none of these answers ;-;
Step-by-step explanation:
3^2 + 8 ÷ 2 - (4 + 3)
we do what's in parentheses first. (pemdas)
so now we have, 3^2 + 8 ÷ 2 - 7
next exponents, 3^2=9
9 + 8 ÷ 2 - 7
next division, 8÷2=4
9 + 4 - 7
next addition, 9+4=13
13-7
finally subtraction, 13-7=6
final answer: 6
Answer:
y = root under 24 (evaluate it if necessary)
or y = 2 root 6
Step-by-step explanation:
Let the reference angle be x
for the triangle in left,
b = 6-4 = 2
Now,
taking x as refrence angle,
cosx = b/h
or, cosx = 2/h
again,
for the bigger triangle,
taking x as reference angle,
cosx = b/h
or, cosx = b/6
As we can see base of bigger triangle is equal to hypotenuse of triangle at the left,
Let's suppose its a
so, cosx = a/6 = 2/a
now,
a/6 = 2/a
or, a² = 12
now,
for bigger triangle, using pythagoras theorem,
h² = p²+b²
or, 6² = y² + a²
or, 36 = y² + 12
or, y² = 24
so, y = root under 24
