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blsea [12.9K]
3 years ago
14

5) The cost of two cakes is $5,what is the cost of nine cakes?

Mathematics
2 answers:
Zinaida [17]3 years ago
4 0

Answer:

I think it’s 22.50 not sure tho

Step-by-step explanation:

If 2 = $5 then u divide 5 by 2 to find the per cost and then multiply with 9 cakes

5/2 = 2.5(9) = 22.5

-Dominant- [34]3 years ago
4 0
It is 22.5 5/2 equals 2.25 2.25 times 9 equals 22.5
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Subtracting this from the numerator gives a remainder of

(x^4+x^3+7x^2-6x+8)-(x^4+2x^3+8x^2)=-x^3-x^2-6x+8

-x^3=-x\cdot x^2. Multiplying the denominator by -x gives

-x(x^2+2x+8)=-x^3-2x^2-8x

and subtracting this from the previous remainder gives a new remainder of

(-x^3-x^2-6x+8)-(-x^3-2x^2-8x)=x^2+2x+8

This last remainder is exactly the same as the denominator, so x^2+2x+8 divides through it exactly and leaves us with 1.

What we showed here is that

\dfrac{x^4+x^3+7x^2-6x+8}{x^2+2x+8}=x^2-\dfrac{x^3+x^2+6x-8}{x^2+2x+8}

=x^2-x+\dfrac{x^2+2x+8}{x^2+2x+8}

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and this last expression is the quotient.

To verify this solution, we can simply multiply this by the original denominator:

(x^2+2x+8)(x^2-x+1)=x^2(x^2-x+1)+2x(x^2-x+1)+8(x^2-x+1)

=(x^4-x^3+x^2)+(2x^3-2x^2+2x)+(8x^2-8x+8)

=x^4+x^3+7x^2-6x+8

which matches the original numerator.

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Answer:

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L = k*cos^{2}(\theta)   (1)

(a) We can rewrite the above equation in terms of sine function using the following trigonometric identity:

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Hence, L reaches its maximum value when θ = 0 because cos²(0) = 1.

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