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4 years ago

Factor the expression below x^2 - 10x + 25

2 answers:

6 0

Find factors of x² and 25 that, when added together, will = -10x

x² - 10x + 25
x -5
x -5

(x - 5)(x - 5) = 0 should be your answer

x - 5 = 0
x - 5 (+5) = 0 (+5)
x = 5

If you are looking for x, x = 5

hope this helps

ivolga24 [154]4 years ago

5 0

It would be (x-5)(x-5)

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Compute i^1+i^2+i^3....i^99+i^100

fgiga [73]

Good morning ☕️

Answer:

Step-by-step explanation:

Consider the sum S = i¹ + i² + i³ +. . .+ i⁹⁹ + i¹⁰⁰

S = i¹ + i² + i³ + . . . + i⁹⁹ + i¹⁰⁰

S = a₁ + a₂ + a₃ +. . . + a₉₉ + a₁₀₀

then, S is the sum of 100 consecutive terms of a geometric sequence (an)

where the first term a1 = i¹ = i and the common ratio = i

FORMULA:______________________

$S=(term1)*\frac{1-(common.ratio)^{number.of.terms}}{1-(common.ratio)}$

_______________________________

then

$S=i*\frac{1-i^{100} }{1-i}$

or i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1 (we know that i⁴ = 1)

Hence

S = 0

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3 years ago

Zane spent $1.50 more than Taylor

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How much does Taylor has?

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3 years ago

* The nth term of sequence A is 3n − 2 The nth term of sequence B is 10 − 2n Sally says there is only one number that is in both

Nina [5.8K]

Answer:

Sally is not right

Step-by-step explanation:

Given the two sequences which have their respective $n^{th}$ terms as following:

Sequence A. $3n - 2$

Sequence B. $10 - 2n$

As per Sally, there exists only one number which is in both the sequences.

To find:

Whether Sally is correct or not.

Solution:

For Sally to be correct, we need to put the $n^{th}$ terms of the respective sequences as equal and let us verify that.

$3n-2=10-2n\\\Rightarrow 3n+2n=10+2\\\Rightarrow 5n=12\\\Rightarrow n = \dfrac{12}{5}$

When we talk about $n^{th}$ terms, $n$ here is a whole number not a fractional number.

But as per the statement as stated by Sally $n$ is a fractional number, only then the two sequences can have a number which is in the both sequences.

Therefore, no number can be in both the sequences A and B.

Hence, Sally is not right.

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3 years ago

Ricky can ride 17 km on his bike in the same length of time ha can walk 9 km.if his riding speed is 4 kph faster than his walkin

aivan3 [116]

Distance=speed times time
if they take the same time, we cal the time , t

b=bike speed
w=walking speed

bikedistnace=17=bt
walkdistance=9=wt

b is 4 more than w
b=4+w

we have
17=bt
9=wt
b=4w

ok so
multiply first equation by 9 and 2nd by 17

153=9bt
153=17wt
set equal
9bt=17wt
divide both sides by t
9b=17w
sub b=4+w for b
9(4+w)=17w
distribute
36+9w=17w
minus 9w both sides
36=8w
divide both sides by 8
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he walks 4.5kph

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3 years ago

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zavuch27 [327]

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Step-by-step explanation:

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3 years ago