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Anarel [89]
2 years ago
14

Determine the median of the following numbers: 40,49,62,56,68,39,50,61,54,44

Mathematics
2 answers:
ExtremeBDS [4]2 years ago
5 0
39 40 44 49 50 54 56 61 62 68
50+54=104/2= 52
Answer is 52
Anit [1.1K]2 years ago
4 0

Answer: 52

Step-by-step explanation: first put all numbers in  numerical order

then find the two centermost numbers

then add those numbers up

then divide the answer by 2

your answer is 52

hope this helps mark me brainliest if it helps

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Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.
Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0
3 years ago
Fine length of BC on the following photo.
MrMuchimi

Answer:

BC=4\sqrt{5}\ units

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

AC^2=AD^2+DC^2

substitute the given values

AC^2=16^2+8^2

AC^2=320

AC=\sqrt{320}\ units

simplify

AC=8\sqrt{5}\ units

step 2

In the right triangle ACD

Find the cosine of angle CAD

cos(\angle CAD)=\frac{AD}{AC}

substitute the given values

cos(\angle CAD)=\frac{16}{8\sqrt{5}}

cos(\angle CAD)=\frac{2}{\sqrt{5}} ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

cos(\angle BAC)=\frac{AC}{AB}

substitute the given values

cos(\angle BAC)=\frac{8\sqrt{5}}{16+x} ----> equation B

step 4

Find the value of x

In this problem

\angle CAD=\angle BAC ----> is the same angle

so

equate equation A and equation B

\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}

solve for x

Multiply in cross

(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units

DB=4\ units

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

BC^2=DC^2+DB^2

substitute the given values

BC^2=8^2+4^2

BC^2=80

BC=\sqrt{80}\ units

simplify

BC=4\sqrt{5}\ units

7 0
2 years ago
Which equation has no solution?
quester [9]

Answer:

all equation has a solution

4 0
2 years ago
Mnv<br><img src="https://tex.z-dn.net/?f=2%20%5Ctimes%20%5Cfrac%7B7%7D%7B%3F%7D%20%20%3D%204" id="TexFormula1" title="2 \times \
faltersainse [42]

Answer:

? = 7/2

Step-by-step explanation:

? = x

a denominator can’t be equal to 0

x ≠ 0

14/x = 4

14 = 4x

x = 7/2

6 0
2 years ago
An expression is shown.<br> 95 – 3^4<br> What is the value of the expression?
Gwar [14]

Answer: 14

Step-by-step explanation:

95-3^4

=95-81

=14

7 0
2 years ago
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