|a| = b gives
a =b or a = -b
so,
|-x| = -10
gives
-x = -10 or -x = 10
x = 10, -10
now let us verify,
when x = 10, |-10| = +10 and it is not = -10
so, x= 10 is NOT a solution.
when x = -10, |-(-10)| = |10| = 10 and it is not = -10
so, x= -10 is NOT a solution.
hence, this equation does not have a solution.
If we know that |...| can never be negative, we can directly deduce that this equation does not have any solution.
Step-by-step explanation:
I believe it's 16 1/2. I just used MathPapa lol.
Answer:
44, 46, and 48
Step-by-step explanation:
add them and you should get 138.
Answer:
x
8
−
256
Rewrite
x
8
as
(
x
4
)
2
.
(
x
4
)
2
−
256
Rewrite
256
as
16
2
.
(
x
4
)
2
−
16
2
Since both terms are perfect squares, factor using the difference of squares formula,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
where
a
=
x
4
and
b
=
16
.
(
x
4
+
16
)
(
x
4
−
16
)
Simplify.
Tap for more steps...
(
x
4
+
16
)
(
x
2
+
4
)
(
x
+
2
)
(
x
−
2
)
Step-by-step explanation:
Given:
The function is:
![f(x)=x^2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2)
The graphs of functions
and
.
To find:
The function
.
Solution:
We have,
![f(x)=x^2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2)
The function
is vertically compresses to get the graph of the function
. So, the function
is:
![g(x)=kf(x)](https://tex.z-dn.net/?f=g%28x%29%3Dkf%28x%29)
...(i)
From the given graph it is clear that the graph of the function
passes through the point (3,3). So, putting
and
in the above function, we get
![3=k(3)^2](https://tex.z-dn.net/?f=3%3Dk%283%29%5E2)
![3=9k](https://tex.z-dn.net/?f=3%3D9k)
![\dfrac{3}{9}=k](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B9%7D%3Dk)
![\dfrac{1}{3}=k](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B3%7D%3Dk)
Putting
in (i), we get
Therefore, the correct option is D.