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2 years ago

What is the vertical distance between (-1,-3) and (-4,-12)?

Mathematics

1 answer:

sasho [114]2 years ago

8 0

Answer:

Use the distance formula to determine the distance between two points.

Exact Form:

3 √ 10

Decimal Form:

9.48683298 …

Step-by-step explanation:

Hope this helps! Have a great day!

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Calculate the limit values:

Nataliya [291]

A) This particular limit is of the indeterminate form,
$\frac{ \infty }{ \infty }$
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }$
So we take the derivatives and obtain,

$Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}$

Still it is of the same indeterminate form, so we apply the rule again,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}$

This simplifies to,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0$

b) This limit is also of the indeterminate form,

$\frac{0}{0}$
we still apply the L'Hopital's Rule,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }$

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }$

When we plug in zero now we obtain,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1$
c) This also in the same indeterminate form

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }$

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }$

It is still of that indeterminate form so we apply the rule again, to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }$

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }$

This gives us;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2$

d) $Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x$

For this kind of question we need to rationalize the radical function, to obtain;

$Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}$

We now divide both the numerator and denominator by x, to obtain,

$Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}$

This simplifies to,

$=\frac{2}{\sqrt{1+0}+1}=1$

5 0

3 years ago

If you could show as much work as possible that would be amazing!!

Nikitich [7]

Answers:

Reduce:

Here we gave to simplify the expressions:

9) $x^{2}+7x+\frac{12}{x^{2}}+11x+28$

Grouping similar terms:

$(x^{2}+\frac{12}{x^{2}})+(18x+28)$

Applying common factor $x^{2}$ in the first parenthesis and common factor $2$ in the second parenthesis:

$x^{2}(1+\frac{12}{x^{4}})+2(9x+14)$ This is the answer

11) $y^{3}+\frac{27}{y^{2}}+2y-3$

Rearranging the terms:

$(y^{3}+2y)+(\frac{27}{y^{2}}-3)$

Applying common factor $y$ in the first parenthesis and common factor $3$ in the second parenthesis:

$y(y^{2}+2)+3(\frac{9}{y^{2}}-1)$ This is the answer

Multiply:

19) $(\frac{12a^{9}u^{7}}{15 c})(\frac{3c^{4}}{21a^{13 u^{8}}})$

Multiplying both fractions:

$\frac{36 a^{9}u^{7}c^{4}}{315 c a^{13}u^{8}}$

Dividing numerator and denominator by 3 and simplifying:

$\frac{12 c^{3}}{105 c a^{4}u}$ This is the answer

21) $(x-\frac{3}{x}-7)(x^{2}-9x+\frac{35}{x^{2}}-18)$

$(\frac{x^{2}-3-7x}{x})(\frac{x^{4}-9x^{3}+35-18x^{2}}{x^{2}})$

Operating with cross product:

$x^{2} (x^{2}-3-7x) x(x^{4}-9x^{3}+35-18x^{2})$

$x^{9} -9x^{8} -18x^{7}+35x^{5}-7x^{8} +63x^{7}+126x^{6}-245x^{4}-3x^{7}+27x^{6}+54x^{5}-105x^{3}$

Grouping similar terms and factoring:

$x^{9}-2(8x^{8}+21x^{7} )+153x^{6}+89x^{5}-5(49x^{4}+21x^{3})$ This is the answer

Divide:

29) $\frac{\frac{k^{6}}{x^{2}}}{\frac{2k^{4}}{3x^{6}}}$

$\frac{3k^{6}x^{6}}{2x^{2}k^{4}}$

Simplifying:

$\frac{3}{2} k^{2}x^{4}$ This is the answer

33) $\frac{\frac{x+5}{x+1}}{\frac{x^{2}+11x+30}{x^{2}+3x+2}}$

$\frac{(x+5)(x^{2}+3x+2)}{(x+1)(x^{2}+11x+30)}$

Factoring numerator and denominator:

$\frac{(x+5)(x+2)(x+1)}{(x+1)(x+6)(x+5)}$

Simplifying:

$\frac{x+2}{x+6}$ This is the answer

37) $\frac{\frac{x-10}{x+13}}{\frac{x^{3}-1000}{x^{2}+15x+21}}$

$\frac{(x-10)(x^{2}+15x+21)}{(x+13)(x^{3}-1000)}$

Applying the distributive property in numerator and denominator:

$\frac{x^{3}+15x^{2}+21x-10x^{2}-150x-210}{(x+13)(x^{4}-1000x+13x^{3}-13000)}$

Grouping similar terms and factoring by common factor:

$-\frac{5x(x^{2}-129)(x^{2}-42)}{1000x^{3} (x+13)(x-13)}$

Dividing by $5$ in numerator and denominator and simplifying:

$-\frac{(x^{2}-129)(x^{2}-42)}{200x^{2}(x+13)(x-13)}$ This is the answer

3 0

3 years ago

2 cm green square is painted on a triangle with a base of 6 and a height of 7. A dot will be randomly placed on the triangle. Wh

agasfer [191]

The area of the triangle is
(1/2)*(6)*(7) = 21 cm²

The area of the square is
(2)(2) = 4 cm²

The area of the triangle that is not covered by the square is
21 - 4 = 17 cm²

The probability that the dot will be placed outside the square is
17/21

Answer: 17/21 (or 0.81)

7 0

2 years ago

The graph of h(x) is a translation of f(x)=3 √x

Gnoma [55]

Answer:

$h(x)=\sqrt[3]{x+2}$