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4 years ago

What is the solution to the equation 9(x − 2) = 27

Mathematics

1 answer:

ehidna [41]4 years ago

4 0

9(x − 2) = 27

x=5

5 is your final answer

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Combine the like terms for the following equations:

ololo11 [35]

1) 15x-24

2) 3b^2

3) m^2-2m-5

4) 14p+7

Hope this helps

mark me as brainiest pls

5 0

3 years ago

Can someone help me find the fa tor of (128m^2n^5 - 200m^4n^3)? <br>

Leona [35]

Answer:

$8m^2n^3$ , $4n-5m$ , and $4n+5m$

Step-by-step explanation:

$128m^2n^5-200m^4n^3$ <-- Given

$m^2(128n^5-200m^2n^3)$ <-- Factor out m²

$m^2n^3(128n^2-200m^2)$ <-- Factor out n³

$8m^2n^3(16n^2-25m^2)$ <-- Factor out 8

$8m^2n^3(4n-5m)(4n+5m)$ <-- Difference of Squares

Therefore, the factors are $8m^2n^3$ , $4n-5m$ , and $4n+5m$

3 0

3 years ago

What is x-y=8 in slope intercept form

iris [78.8K]

Answer:

y = x-8

Step-by-step explanation:

Slope intercept form is

y = mx+b where m is the slope and b is the y intercept

x-y = 8

Subtract x from each side

x-y-x = -x+8

-y = -x+8

Divide each side by -1

-y/-1 = -x/-1 +8/-1

y = x-8

5 0

4 years ago

What is greater 4.58 or 4.563

photoshop1234 [79]

Answer:

4.58 is greater then 4.563

3 0

3 years ago

Read 2 more answers

What will be the value of

madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

Then you want to know the value of

$\displaystyle\lim_{n\to\infty}a_n$

if it exists.

To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.