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3 years ago

Please anybody help me out

Mathematics

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

50 is the awnser

Step-by-step explanation:

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Black_prince [1.1K]

Answer:

CB= 4

Step-by-step explanation:

the triangles are in AA similarity.

AB/AD = BC/DE

4/12 = x/12

CROSS MULTIPLY;

x = 48/12

x = 4

5 0

3 years ago

If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?

enyata [817]

Step-by-step explanation:

$\bf➤ \underline{Given-} \\$

$\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\$

$\bf➤ \underline{To\: find-} \\$

$\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\$

$\bf ➤\underline{Solution-} \\$

Let us assume that:

$\rm: \longmapsto u = {x}^{2013}$

Therefore, the equation becomes:

$\rm: \longmapsto u + \dfrac{1}{u} = 2$

$\rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2$

$\rm: \longmapsto{u}^{2} + 1 = 2u$

$\rm: \longmapsto{u}^{2} - 2u + 1 =0$

$\rm: \longmapsto {(u - 1)}^{2} =0$

$\rm: \longmapsto u = 1$

Now substitute the value of u. We get:

$\rm: \longmapsto {x}^{2013} = 1$

$\rm: \longmapsto x = 1$

Therefore:

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 1 + 1$

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2$

★ Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

6 0

3 years ago

What is .25 written in its simplest fraction form? A. 2/5 B. 4/1 C. 1/4 D. 100/25

Furkat [3]

0.25 = 25/100

to simplify 25/100, divide 25 from both the numerator and denominator.

25/25 = 1

100/25 = 4

1/4 is the simplest form

C. 1/4 is your answer

hope this helps

5 0

3 years ago

Read 2 more answers

Find the sum of the geometric series 512+256+ . . .+4

mario62 [17]

$\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4$

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

$\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}$

$\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}$

so is the 8th term, then, let's find the Sum of the first 8 terms.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}$

$\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020$

7 0

3 years ago

Is y=6x-2 proportional

Alexus [3.1K]

Yes, the equation given above is proportional. The variables y and x are directly proportional to each other. Two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant multiplier.

7 0

3 years ago

Read 2 more answers