Answer:
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
Step-by-step explanation:
The domain is all the x-values of a relation.
The range is all the y-values of a relation.
In this example, we have an equation of a circle.
To find the domain of a relation, think about all the x-values the relation can be. In this example, the x-values of the relation start at the -1 line and end at the 3 line. The same can be said for the range, for the y-values of the relation start at the -4 line and end at the 0 line.
But what should our notation be? There are three ways to notate domain and range.
Inequality notation is the first notation you learn when dealing with problems like these. You would use an inequality to describe the values of x and y.
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
Set-builder notation is VERY similar to inequality notation except for the fact that it has brackets and the variable in question.
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
Interval notation is another way of identifying domain and range. It is the idea of using the number lines of the inequalities of the domain and range, just in algebriac form. Note that [ and ] represent ≤ and ≥, while ( and ) represent < and >.
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
here,
2x-6y= -12
2(x-3y)= -12
x-3y = -6...........eq1
x+2y=14............eq2
we can subtract the above two given equations
we get,
___________
so,
y=4
now to find the value of x we have to substitute the value of y in any of the above two equations, I choose eq2
we get,
<h3>so,</h3><h3> x=6,y=4</h3>
The chemical structure includes the bonding angle, the type of bonds, the size of the molecule, and the interactions between molecules. Slight changes in the chemical structure can drastically affect the properties of the compound.
Answer:
<u>y'= 5x^4 + 5^x In(5)</u>
Step-by-step explanation:
<u>Differentiate</u><u> </u><u>with </u><u>Respect</u><u> </u><u>to</u><u> </u><u>x</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>In(</u><u>5</u><u>^</u><u>x</u><u>)</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>x </u><u>In(</u><u>5</u><u>)</u>
<u>with </u><u>respect</u><u> </u><u>to </u><u>x,</u><u> </u><u>we </u><u>have</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>y </u><u>In(</u><u>5</u><u>)</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>5</u><u>^</u><u>x</u><u> </u><u>In(</u><u>5</u><u>)</u>
