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2 years ago

Grazer sells cool Minecraft t-shirts through his

Mathematics

2 answers:

zmey [24]2 years ago

7 0

Answer:

125

Step-by-step explanation:

125 x7 = 875

notsponge [240]2 years ago

3 0

Answer:

125 T-shirts

Step-by-step explanation:

Divide $875 by $7. There is your answer.

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Which will result in a difference of squares?

otez555 [7]

Expanding the given expressions using Foil:

1)(–7x + 4)(–7x + 4) = $7x^{2}-28x-28x+16=49x^{2}-56x+16$

2) (–7x + 4)(4 – 7x)= $-28x+49x^{2}+16-28x=49x^{2}-56x+16.$

3)(–7x + 4)(–7x – 4)= $49x^{2}+28x-28x-16$ = $49x^{2}-16.$

4)(–7x + 4)(7x – 4)= $49x^{2}+28x+28x-16=49x^{2}+56x-16.$

The third option that is (–7x + 4)(–7x – 4) is difference of two squares.

5 0

3 years ago

Read 2 more answers

Solving two step equations w/-3+14=5

natali 33 [55]

w/–3 + 14 = 5

⇛ w/–3 = 5 – 14

⇛ w/–3 = – 11

⇛ w = –11 × (–3)

⇛ w = 33

7 0

2 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:

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