Quadratic Equations in Single Variable
n²-3n+10=0
2x²+2x+1=0
25b²-16=0
f²-3f+2=0
1/3m +2m=4
a²=225
Linear Equation in single Variable
8-3k=12
5w+5=0
10u-5=8
Linear Equation in Two Variable
2y-z=9
3r+2e=6
d=3e-7
<h3>What is an equation?</h3>
It should be noted that an equation simply means the expression that's used to show the relationship between the variables.
In this case, the equation has been grouped.
Learn more about equations on:
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Group the given equations into two based on observed common properties.
Equations
n²-3n+10=0
8-3k=12
2y-z=9
2x²+2x+1=0
25b²-16=0
3r+2e=6
5w+5=0
f²-3f+2=0
d=3e-7
1/3m +2m=4
10u-5=8
a²=225
The segment of length x bisects the chord of length 19.2.
The diameter is 24, so the radius is 12.
You have a right triangle with hypotenuse 12, and one leg 19.2/2 = 9.6
9.6^2 + x^2 = 12^2
92.16 + x^2 = 144
x^2 = 51.84
x = 7.2
Answer:
<h3>Find the explanation below</h3>
Step-by-step explanation:
A) Given the equation z - 5 = 2
To get the solution, we will add 5 to both sides and find z as shown;
z - 5 = 2
z - 5 + 5 = 2 + 5
z + 0 = 2+5
z = 7
Hence z = 5 is not the solution of the equation but z = 7
B) Given the inequality expression t+2>5
Substract 2 from both sides of the equation;
t+2 - 2>5 - 2
t + 0> 3
t>3
Hence the solution t = 4 is not true for the expression. The solution is t > 3
C) Given the expression x-2=2, to get the solution to the expression, you will add 2 to both sides of the equation;
x - 2+ 2 = 2 + 2
x + 0 = 4
x = 4
Therefore x = 4 is the solution of this equation.
Based on the question, only Option C is correct
Answer:
Step-by-step explanation:
H0 : μ = 46300
H1 : μ > 46300
α = 0.05
df = n - 1 = 45 - 1 = 44
Critical value for a one tailed t-test(since population standard deviation is not given).
Tcritical = 1.30
The test statistic :(xbar - μ) ÷ (s/sqrt(n))
The test statistic, t= (47800-46300) ÷ (5600√45)
t = 1500
t = 1500 / 834.79871
t = 1.797
The decision region :
Reject H0: if t value > critical value
1. 797 > 1.30
Tvalue > critical value ; We reject H0
Hence, there is sufficient evidence to conclude that cost has increased.
