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Spočtěte $\int_{7\pi}^{8\pi}\limits \cos \frac{x+4\pi}{6}\ \mathrm{d}x$ \int_{7\pi}^{8\pi}\limits \cos \frac{x+4\pi}{6}\ \mathrm{d}x

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Spočtete obsah rovinného obrazce ohraničeného křivkami o rovnicích x=2, x=5, y=0 a y=4+(2·x).

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Spočtěte neurčitý integrál

$\displaystyle\int\left(\frac{1}{x^{9}}+e\right)dx.$ \displaystyle\int\left(\frac{1}{x^{9}}+e\right)dx.

$\displaystyle\ln\left|x^{9}\right|+e^x+C$ \displaystyle\ln\left|x^{9}\right|+e^x+C

$\displaystyle\frac{-1}{8x^{8}}+e^x+C$ \displaystyle\frac{-1}{8x^{8}}+e^x+C

$\displaystyle\frac{10}{x^{10}}+e^x+C$ \displaystyle\frac{10}{x^{10}}+e^x+C

$\displaystyle\frac{10}{x^{10}}+ex+C$ \displaystyle\frac{10}{x^{10}}+ex+C

$\displaystyle\ln\left|x^{9}\right|+ex+C$ \displaystyle\ln\left|x^{9}\right|+ex+C

$\displaystyle\frac{-1}{8x^{8}}+ex+C$ \displaystyle\frac{-1}{8x^{8}}+ex+C

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Spočtěte $\displaystyle\int_{0}^{\frac{1}{2}}\frac{2}{\sqrt{1-x^2}}\ \mathrm{d}x$ \displaystyle\int_{0}^{\frac{1}{2}}\frac{2}{\sqrt{1-x^2}}\ \mathrm{d}x (zaokrouhlete na dvě desetinná místa).

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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x-\frac{\pi}{4})+3.$ f(x)=\cos(3x-\frac{\pi}{4})+3. se středem v bodě $a=\frac{\pi}{3}$ a=\frac{\pi}{3}.

$3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 2}{4}(x-\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 2}{4}(x-\frac{\pi}{3})^2

✅

$3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 2}{2}(x-\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 2}{2}(x-\frac{\pi}{3})^2

❌

$3-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{\sqrt 2}{4}(x-\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{\sqrt 2}{4}(x-\frac{\pi}{3})^2

❌

$3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 2}{2}(x-\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 2}{2}(x-\frac{\pi}{3})^2

❌

$3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 2}{4}(x-\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 2}{4}(x-\frac{\pi}{3})^2

❌

$3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x+\frac{\pi}{3})-\frac{9\sqrt 2}{2}(x+\frac{\pi}{3})^2$ 3-\frac{\sqrt 2}{2}-\frac{3\sqrt 2}{2}(x+\frac{\pi}{3})-\frac{9\sqrt 2}{2}(x+\frac{\pi}{3})^2

❌

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Recall the PAE model, where

PAE = C + I^P + G;

C = C₀ + c(Y-T) ; I^P = I₀ ; G= G₀; T = T₀ + tY.

The diagram below depicts this PAE model.

In the diagram above, what can explain the movement from PAE₁to PAE_new?

Decline in G

Decline in C0

All of the above options.

Види контролю і самоконтролю як методів навчання:

усний контроль, письмовий, колоквіум

усний контроль, письмовий.

усний контроль, письмовий, машинний, тестовий;

100%

усний контроль, письмовий, іспит, залік

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With the evolution of the financial system, combined with advancements in monetary policy and risk management, many countries (e.g., Australia and Canada) have abolished some minimum required ______ on banks. This practice can boost economic activity and foster economic growth.

Which of the following ratio is the best fit to fill in the blank?

Reserve-deposit ratio

Liquidity coverage ratios

❌

Net stable funding ratios

❌

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