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garik1379 [7]
4 years ago
12

Which statement most accurately explains why x^5/x^2 is equal to x^3?

Mathematics
2 answers:
Serggg [28]4 years ago
8 0
\bf \cfrac{x^5}{x^2}\implies \cfrac{x^{3+2}}{x^2}\implies \cfrac{x^3\underline{x^2}}{\underline{x^2}}\implies x^3
hodyreva [135]4 years ago
8 0
The answer you are looking for would be C.

When you divide varibles with exponents, you subtract the exponents. Since there are no numbers in front of the x, we can assume there is only 1 x. 1x/1x = 1x. Since the top has the largest exponent, you'd subtract the bottom exponent to the top, leaving a power of 3. Thus meaning, the answer of x^3 is explained in C.

I hope this helps!
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Find the value of x and y. (geometry)
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36+36=72
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3 years ago
Anna picked 6 apples. Sabrina picked 7 times as many apples. How many apples did Sabrina pick?
koban [17]

Anna picked 6 apples

Sabrina picked 7 <em>times</em> as many apples.

Multiply the two numbers together

6 x 7 = 42

Sabrina picked 42 apples

hope this helps

~<em>Rise Above the Ordinary</em>

6 0
4 years ago
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Calculate y as a function of x when dy/dx = 4x3 + 3x2 - 6x + 5
PtichkaEL [24]
Answer:
y = x⁴ + x³ - 3x² + 5x + C

======

Separable differential equations such as these ones can be solved by treating dy/dx as a ratio of differentials. Then move the dx with all the x terms and move the dy with all the y terms. After that, integrate both sides of the equation.

   \begin{aligned}&#10;\dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\&#10;dy &= (4x^3 + 3x^2 - 6x + 5) dx \\&#10;\int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx &#10;\end{aligned}

In general (understood that +C portions are still there), 

   \int x^{m} = \dfrac{x^{m+1}}{m+1}

Note that ∫dy = y  since it is ∫1·dy = ∫y⁰ dy = y¹/(0+1) = y
For the right-hand side, we use the sum/difference rule for integrals, which says that

   \int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx

Applying these concepts:

   \begin{aligned} &#10; \int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\&#10;y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\&#10;&= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one $C$ is needed)} \\&#10;&= x^4 + x^3 - 3x^2 + 5x + C&#10;\end{aligned}

The answer is y = x⁴ + x³ - 3x² + 5x + C
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Hitman42 [59]

Answer:

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