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4 years ago

Let f(x) = 27x5 – 33x4 – 21x3 and g(x) = 3x2. Find f(x)/g(x)

Mathematics

1 answer:

marissa [1.9K]4 years ago

8 0

(27x^5 - 33x^4 - 21x^3)/3x^2 = (27/3)x^(5 - 2) - (33/3)x^(4 - 2) - (21/3)x^(3 - 2) = 9x^3 - 11x^2 - 7x

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In 1 hour the employee dropped 635 chocolates. If this represents 25% of the days chocolate molds, how many chocolates were plan

Zinaida [17]

Answer:

2540

Step-by-step explanation:

25% is a quarter of 100% so if 635 is 25% you multiple 635 x 4 to get your final answer concluding in 2540. if i'm wrong let someone else correct me :b

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3 years ago

Write log7(2 ⋅ 6) + log73 as a single log. Log Subscript 7 Baseline 11 Log Subscript 7 Baseline 15 Log Subscript 7 Baseline 36

zhannawk [14.2K]

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3 years ago

Read 2 more answers

Given a * b=ba−ba+ab, find (2*3)×(3*2).

Vladimir79 [104]

Answer:

Step-by-step explanation:

a = (2×3) = 6

b = (3×2) = 6

Substitue into the equation:

a×b=ba−ba+ab

= 6×6 - 6×6 + 6×6

= 36 - 36 + 36

= 36

Therefore, (2*3)×(3*2) = 36

Hope this helps.☺

4 0

3 years ago

In a rectangle, the sides have measurements of 2x + 1 and 3x - 5 units Find an expression for the area of the rectangle, in squa

SashulF [63]

Answer:

$(6 {x}^{2} - 7x - 5) \: {units}^{2}$

Step-by-step explanation:

Area of rectangle = Product of sides

$= (2x + 1) \times (3x - 5) \\ = 2x(3x - 5) + 1(3x - 5) \\ = 6 {x}^{2} - 10x + 3x - 5 \\ area \: of \: rectangle \: \\ = (6 {x}^{2} - 7x - 5) \: {units}^{2}$

4 0

4 years ago

How can I evaluate this question?

ohaa [14]

$\bf \left(x^2-\cfrac{2}{\sqrt{x}}+1 \right)(\sqrt[3]{x}+3x-4)\quad 
\begin{cases}
\frac{2}{\sqrt{x}}\implies \frac{2}{x^{\frac{1}{2}}}\implies 2x^{-\frac{1}{2}}\\\\
\sqrt[3]{x}\implies x^{\frac{1}{3}}
\end{cases}
\\\\\\
(x^2-2x^{-\frac{1}{2}}+1)(x^{\frac{1}{3}}+3x-4)$

$\bf \\\\\\
\begin{cases}
x^2\cdot x^{\frac{1}{3}}+3x^3-4x^2\\\\
-2x^{-\frac{1}{2}}\cdot x^{\frac{1}{3}}-2x^{-\frac{1}{2}}\cdot 3x+2x^{-\frac{1}{2}}\cdot 4\\\\
+x^{\frac{1}{3}}+3x-4
\end{cases}
\\\\\\ 
\begin{cases}
x^{2+\frac{1}{3}}+3x^3-4x^2\\\\
-2x^{-\frac{1}{2}+\frac{1}{3}}-6x^{-\frac{1}{2}+1}+8x^{-\frac{1}{2}}\\\\
+x^{\frac{1}{3}}+3x-4
\end{cases}$

$\bf x^{\frac{7}{3}}+3x^3-4x^2-2x^{-\frac{1}{6}}-6x^{\frac{1}{2}}+8x^{-\frac{1}{2}}+x^{\frac{1}{3}}+3x-4
\\\\\\
\sqrt[3]{x^7}+3x^3-4x^2-\cfrac{2}{x^{\frac{1}{6}}}-6\sqrt{x}+\cfrac{8}{x^{\frac{1}{2}}}+\sqrt[3]{x}+3x-4
\\\\\\
x^2\sqrt[3]{x}+3x^3-4x^2-\cfrac{2}{\sqrt[6]{x}}-6\sqrt{x}+\cfrac{8}{\sqrt{x}}+\sqrt[3]{x}+3x-4$

8 0

3 years ago