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3 years ago

May someone please help with the way please

Mathematics

1 answer:

Oksana_A [137]3 years ago

5 0

Answer:

17.6 m²

Step-by-step explanation:

Given the ratio of similar shapes = a : b, then

area of shapes = a ² : b²

Δ PTQ and Δ PRS are similar and so the ratio of corresponding sides are equal, that is

PT : PR = 6 : 9 = 2 : 3, thus

ratio of areas = 2² : 3² = 4 : 9

let the area of Δ PQT be x, then using proportion

$\frac{4}{x}$ = $\frac{9}{x+22}$ ( cross- multiply )

9x = 4(x + 22) ← distribute

9x = 4x + 88 ( subtract 4x from both sides )

5x = 88 ( divide both sides by 5 )

x = 17.6

Thus area of Δ PQT = 17.6 m²

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julsineya [31]

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$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

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We can of course see the limits are identical by replacing $x\mapsto e^x$ , so that

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You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$

so the overall limit is

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2 years ago

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Answer:

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3 years ago

Find the distance from point A(−14, 5) to the line −x+2y = 14.

Ad libitum [116K]

Answer:

The distance from point A to given line is: 2√5 or 4.472 units

Step-by-step explanation:

Given equation of line is:

$-x+2y= 14$

And the point is:

$A = (x_1,y_1) = (-14,5)$

The standard form of equation of line is:

$Ax+By+C = 0$

Converting the given equation into standard form

$-x+2y-14=0$

The distance of a point (x1,y1) from a line is given by the formula:

$d = \frac{|Ax_1+By_1+C}{\sqrt{A^2+B^2}}$

In the given details,

A = -1, B = 2, C = -14

x1 = -14, y1 = 5

Putting the values in the formula

$d = \frac{|(-1)(-14)+(2)(5)-14|}{\sqrt{(-1)^2+(2)^2}}\\d = \frac{14+10-14}{\sqrt{1+4}}\\d = \frac{10}{\sqrt{5}}\\d = \frac{5*2}{\sqrt{5}}\\d = \frac{\sqrt{5}*\sqrt{5}*2}{\sqrt{5}}\\d = 2\sqrt{5}\\d = 4.472\ units$

Hence,

The distance from point A to given line is: 2√5 or 4.472 units

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Vadim26 [7]

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