The domain is the set of all real numbers and the range is the set of all real numbers larger than -4, so the correct option is the third one.
<h3 /><h3>How to get the domain and the range?</h3>
For a function y = f(x) we define the domain as the set of the x-values (horizontal axis) and the range as the set of the y-values (vertical axis).
Here we have a parabola, in this case the domain is always the set of all real numbers, and the range is all the values above or equal to the vertex (in cases like this, where the parabola opens up).
We can see that the vertex is at y = -4
So the range is.
[-4, ∞]
So the correct option is 3.
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The first step to solving this is to factor out the perfect square of 5²

next,, factor out the perfect square of p³

the root of a product is equal to the product of the roots of each factor
√5² √p² √p
now reduce the first index of the radical and the exponent with 2
5√p² √p
then reduce the second index of the radical and exponent with
5p√p
this means that the correct answer to your question is 5p√p
let me know if you have any further questions
:)
C = 169
square is (x + 13)^2
Explanation:
In the form:
ax^2+ bx +c
to complete the square a = 1 is required.
c = (b/2)^2
the square term is b/2
x^2 + 26x + c
a = 1 so we are okay to complete the square.
b = 26
c = (26/2)^2 = 13^2 = 169
finally to complete the square:
x^2 + 26x + 169 = (x+b/2)^2 = (x+13)^2
Answer:
Multiply the coordinate by the scale factor, 1/5 to get (7/25,-5/20). You can simplify this to (7/25, -1/4)
Answer:
0.10865 killograms
Step-by-step explanation:
calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.
Given:
temperature: 
quality : 50% = 0.5
initial mass: 
to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.
by taking the difference of mass we can determine the mass of liquid removed.
we have two phases at temperature
with specific volume for liquid
and specific volume for vapor is
.
The Volume in the initial state is given by, (Using definition of specific volume)

using 

substituting
,
, 
we get
finally
we know the formula to find liquid mass is

density of ammonia is 
inserting the values into the formula we get the value for liquid mass removed through the valve.

the final answer is
