How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove
1 answer:
The equation has one extraneous solution which is n ≈ 2.38450287.
Given that,
The equation;
We have to find,
How many extraneous solutions does the equation?
According to the question,
An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.
To solve the equation cross multiplication process is applied following all the steps given below.
The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y
with 0 and solve for x. The graph of the equation is attached.
n ≈ 2.38450287
Hence, The equation has one extraneous solution which is n ≈ 2.38450287
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Answer:
The cooking club sales covers the expenditure when 2 piece of cakes are sold.
Step-by-step explanation:
Given:
Selling price of each piece of cake = $10
Cost for booth at fair = $10
Ingredients for each piece of cake = $5
We need to find the number of pieces of cake sold when the sales cover the expenditures.
Solution:
Let the number of pieces be 'x'
So We can say that the point at which the sales cover the expenditures can be calculated as Selling price of each piece of cake multiplied by number of pieces will be equal to Cost for booth at fair plus Ingredients for each piece of cake multiplied number of piece of cakes.
framing in equation form we get;
Now Subtracting both side by '5x' using Subtraction property we get;
Now Dividing both side by 5 we get;
Hence The cooking club sales covers the expenditure when 2 piece of cakes are sold.
Answer:
Step-by-step explanation:
1. Given the integral function , using trigonometric substitution, the substitution that will be most helpful in this case is substituting x as
i.e
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All integrals in the form are always evaluated using the substitute given where 'a' is any constant.
From the given integral, where a = 7 in this case.
The substitute will therefore be
2.) Given
cross multiplying
3.) Rewriting the given integral using the substiution will result into;