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3 years ago

Assuming these triangles are similar,solve for x.

Mathematics

2 answers:

alexgriva [62]3 years ago

7 0

Answer:

x = 7

Step-by-step explanation:

40/16 = 2.5 scale factor that it's being enlarged by. So, 16 x 2.5 =40

45/2.5 = 18

2x + 4 = 18

- 4 both side

2x = 14

÷2 both side

x = 7

UNO [17]3 years ago

3 0

Answer:

Step-by-step explanation:

all the difference between these polygons is that the left one is 2.5x bigger than the one on the right (I found this out by doing 40/16 which is 2.5)

so now I need to divide 45 by 2.5

45/2.5=18

Now I know that

2x+4=18

-4 from both sides

2x=14

divide by 2

x=7

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Butoxors [25]

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A=Pe^{rt}\qquad 
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A=\textit{accumulated amount}\to &90,000\\
P=\textit{original amount}\to& \$65,452\\
r=rate\to 9.1\%\to \frac{9.1}{100}\to &0.091\\
t=years
\end{cases}$

$\bf 90000=65452e^{0.091t}\implies \cfrac{90000}{65452}=e^{0.091t}\implies \cfrac{22500}{16363}=e^{0.091t}
\\\\\\
\textit{taking \underline{ln} to both sides}\qquad ln\left( \cfrac{22500}{16363} \right)=ln\left( e^{0.091t} \right)
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ln\left( \cfrac{22500}{16363} \right)=0.091t\implies 
\cfrac{ln\left( \frac{22500}{16363} \right)}{0.091}=t$

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4 years ago

Simplifying Fractions<br><br> 10/45 simplified? Thanks!<br> Also, 20 points c;

vichka [17]

10/45 simplified is 2/9. Since both numbers are divisible by 5. Hope i helped you today!

10 ÷5 2
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yanalaym [24]

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Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

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3 years ago