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sergiy2304 [10]

3 years ago

14

Can you please help me i need to factor this equation but i am stuck in this type of question: c^2-9cd+18d^2

1 answer:

spin [16.1K]3 years ago

4 0

C^2-3cd-6cd+18d^2
=c(c-3d) -6d(c-3d)
=(c-3d) (c-6d)

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Kena calculates that she spends 15% of a school day in science class. If she spends 75 minutes in science class, how many minute

ankoles [38]

Answer:

Step-by-step explanation:

5 0

3 years ago

2w-5=25<br> 10<br> 15<br> -15<br> -10<br> No solution <br> All real number

mihalych1998 [28]

Answer:

w=15

Step-by-step explanation:

2w-5=30

2×15-5=30

30-5=25

w=15

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3 years ago

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Which of the following illustrate the commutative property? select all that apply

tatiyna

The answer choices would be: A, D, and E. These would be the correct answer choices because commutative property only works for addition and multiplication. Therefore, C and F are incorrect. B would also be incorrect because it is showing the Distributive Property. I hope this helps!

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3 years ago

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Ms. Check your answers.

xenn [34]

3kg+2kg=5kg
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5 0

3 years ago

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Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

4 years ago