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4 years ago

Why isint 11/12 equivalent to 3/4? i need a simple answer that i can write on my work

Mathematics

2 answers:

Bad White [126]4 years ago

8 0

If $\frac{11}{12}= \frac{3}{4}$ ⇒3*12=11*4 ⇒ 36=44 false
⇒ 11/12 isn't equal with 3/4

rodikova [14]4 years ago

7 0

12 / 4 = 3. That means that 3 lots of 4 go into 12. If the 2 fractions are equivalent, then the numerator (the top number) for 3/4 times by 3 should equal the numerator for 11/12. However, 3 X 3 = 9. 9/12 is not the same as 11/12.

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Veena ate lunch at a deli. She ordered a turkey snadwich for $9.25 and a salad for 4.35. The tax was $7.5. What is the amount of

Nastasia [14]

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4 years ago

Read 2 more answers

Sarah will buy bracelets and necklaces in order to sell them in her schools fund raising activity. The bracelets cost $2 and the

Alborosie

Answer:

The answer will be:

b ≥ 10

Step-by-step explanation:

The bracelets cost $2 and the earrings cost $3.

She needs to make at least $60.

She will need to sell more than 10 bracelets.

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6 0

3 years ago

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$