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Salsk061 [2.6K]
3 years ago
11

How to work out 20% of £54

Mathematics
1 answer:
Alexandra [31]3 years ago
3 0

Answer:

£10.80

Step-by-step explanation:

20% = \frac{20}{100} = 0.2 , then

£54 × 0.2 = £10.80

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Match each quadratic function to its graph.
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Answer:

You have to have the graph to answer it.

Step-by-step explanation:

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Which relation below represents a one to one function
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Answer:

<h2>Second table</h2>

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<em>One-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. Every element of the function's domain is the image of at most one element of its domain.</em>

First table:

No. Because for x = 7.25 and x = 8.5 we have the same value of y = 11

Second table:

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Please help I’ve been struggling for ages!
Lapatulllka [165]

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Answer:

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Step-by-step explanation:

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8 0
3 years ago
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
seropon [69]

Answer:

h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}

3) To test it, just integrate:

\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C

5 0
3 years ago
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