Answer:
x = 56/3
PS = 103/3
Step-by-step explanation:
Perimeter: (4x + 1) + 2(2x - 3) + (4 + 4 + 4x + 1) = 228
Collect and combine like terms: 12x + 4 = 228
Subtract 4 from both sides: 12x = 224
Divide both sides by 12: x = 56/3
PS = (2 x 56/3) - 3 = 103/3
<span>We have 2 given numbers:
=> 86 + 68 + 38 = 192
=> 68 + 38 + 86 = 192 also
This kind of process used commutative property system of equation in where:
a + b = b + a
No matter what sequence you are going to use, the answer would always be the
same.
In the given 2 equations, notice that both digits of 2 group of equation
contains the same number. The only difference is they were added with different
orders. Yet the answers are still the same.
</span>
The entire range of independent variable values is the domain of a function.
After substituting the domain, the range of just a function is the entire set of all possible values for the dependent variable (often y).
What is domain and range?
- The collection of all x-values that can cause the function to "work" and produce actual y-values is known as the domain.
- The range is the set of y-values that are produced when all the conceivable x-values are substituted.
The entire range of independent variable values is the domain of a function.
Keep these things in mind when locating the domain:
- A fraction's denominator (bottom) cannot be 0.
- In this section, the integer following a square root symbol must be positive.
After substituting the domain, the range of just a function is the entire set of all possible values for the dependent variable (often y).
The variety of potential y-values makes up a function's range (minimum y-value to maximum y-value)
- To observe what happens, substitute several x-values into the expression for y.
- Be sure to search for the least and highest y values.
Learn more about Domain and Range here:
brainly.com/question/10197594
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The answer is 31/35
Hope it helps!
The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.
