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2 years ago

SWYK: Triangle Midsegment

Mathematics

1 answer:

Ganezh [65]2 years ago

8 0

Answer:

Step-by-step explanation:

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What does n/5 = ? <br><br> A) f/9<br><br> B) 9/5

nadezda [96]

Answer:

I think is A but check it just in case

Step-by-step explanation:

3 0

3 years ago

Simplify 5 over -32 squared simplified

Brilliant_brown [7]

Answer:

Step-by-step explanation:

−322

−1024

−5

1024

8 0

3 years ago

Read 2 more answers

At a meeting with 25 businessmen, twelve businessmen drank coffee, and ten businessmen drank tea. Five businessmen drank both co

likoan [24]

Answer:

Step-by-step explanation:

25 total people

5 drank coffee and tea

tea drinkers only : 10 - 5 = 5

coffee drinkers only : 12 - 5 = 7

total people who drank anything : 7 + 5 + 5 = 17

number of people who drank neither : 25 - 17 = 8 <=====

4 0

3 years ago

40, 10, 5/2 5/8... (fractions) is it arithmetic or geometric and what are the next two terms PLZ HELP DUE IN 10 MINS

patriot [66]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the sequence

$40,\:10,\:\frac{5}{2},\:\frac{5}{8}$

A geometric sequence has a constant ratio 'r' and is defined by

$\:a_n=a_0\cdot r^{n-1}$

Computing the ratios of all the adjacent terms

$\frac{10}{40}=\frac{1}{4},\:\quad \frac{\frac{5}{2}}{10}=\frac{1}{4},\:\quad \frac{\frac{5}{8}}{\frac{5}{2}}=\frac{1}{4}$

The ratio of all the adjacent terms is the same and equal to

$r=\frac{1}{4}$

Thus, the given sequence is a geometric sequence.

As the first element of the sequence is

$a_1=40$

Therefore, the nth term is calculated as

$\:a_n=a_0\cdot r^{n-1}$

$a_n=40\left(\frac{1}{4}\right)^{n-1}$

Put n = 5 to find the next term

$a_5=40\left(\frac{1}{4}\right)^{5-1}$

$a_5=40\cdot \frac{1}{4^4}$

$a_5=\frac{40}{4^4}$

$=\frac{2^3\cdot \:5}{2^8}$

$a_5=\frac{5}{2^5}$

$a_5=\frac{5}{32}$

now, Put n = 6 to find the 6th term

$a_6=40\left(\frac{1}{4}\right)^{6-1}$

$a_6=40\cdot \frac{1}{4^5}$

$a_6=\frac{40}{4^5}$

$=\frac{2^3\cdot \:5}{2^{10}}$

$a_6=\frac{5}{2^7}$

$a_6=\frac{5}{128}$

Thus, the next two terms of the sequence 40, 10, 5/2, 5/8... is:

$a_5=\frac{5}{32}$

$a_6=\frac{5}{128}$

7 0

3 years ago

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

3 years ago