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cluponka [151]
3 years ago
5

Factor: 3x^(2)-16x+16

Mathematics
2 answers:
lana66690 [7]3 years ago
7 0
Answer is (3x-4) (x-4)
This is the answer for Plato and Edmentum
fenix001 [56]3 years ago
3 0

Answer:

3 {x}^{2}  - 16x + 16 \\ 3 {x}^{2}  - 12x - 4x + 16 \\ 3x(x - 4) - 4(x - 4) \\ (3x - 4)(x - 4)

Step-by-step explanation:

I hope it helped U

stay safe stay happy

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Hannah was asked to make d the subject of the formula<br> d-7=4d+3/e<br> Finish her question.
tigry1 [53]

Answer:

d = \frac{-1}{e} - \frac{7}{3}

Step-by-step explanation:

In order to make d the subject of the formula, you need to isolate it.

- You started with d-7 = 4d + 3/e

- Move 4d to the left side by subtracting 4d from both sides to cancel it from the right so you have...

d - 7 = 4d + 3/e              This will leave you left with -3d - 7 = 3/e

-4d     -4d                          

- Then move over the -7 by adding 7 to both sides...

-3d - 7 = 3/e              This will leave you left with -3d = 3/e + 7

     +7          +7

- Finally to get d by itself divide both sides of the equation by -3 and you'll be left with...

d = \frac{3}{-3e} -\frac{7}{3}

- You can cancel out the 3 in the -3/3e and make it -1/e so your final answer will be

d = \frac{-1}{e} - \frac{7}{3}

5 0
3 years ago
What is the missing pattern 7, 11, 2, 18, -7
Alexxx [7]

The <em>missing</em> pattern behind the sequence 7, 11, 2, 18, -7 is described by the formula n = 7 + \sum \limits_{i= 1}^{n} (-1)^{i+1}\cdot (i + 1)^{2}, equivalent to the <em>recurrence</em> formula a_{n+1} = a_{n} + (-1)^{i+1}\cdot (i + 1)^{2}.

<h3>What is the missing element in a sequence?</h3>

A sequence is a set of elements which observes at least a <em>defined</em> rule. In this question we see a sequence which follows this rule:

n = 7 + \sum \limits_{i= 1}^{n} (-1)^{i+1}\cdot (i + 1)^{2}      (1)

Now we prove that given expression contains the pattern:

n = 0

7

n = 1

7 + (- 1)² · 2² = 7 + 4 = 11

n = 2

7 + (- 1)² · 2² + (- 1)³ · 3² = 11 - 9 = 2

n = 3

7 + (- 1)² · 2² + (- 1)³ · 3² + (- 1)⁴ · 4² = 2 + 16 = 18

n = 4

7 + (- 1)² · 2² + (- 1)³ · 3² + (- 1)⁴ · 4² + (- 1)⁵ · 5² = 18 - 25 = - 7

The <em>missing</em> pattern behind the sequence 7, 11, 2, 18, -7 is described by the formula n = 7 + \sum \limits_{i= 1}^{n} (-1)^{i+1}\cdot (i + 1)^{2}, equivalent to the <em>recurrence</em> formula a_{n+1} = a_{n} + (-1)^{i+1}\cdot (i + 1)^{2}.

To learn more on patterns: brainly.com/question/23136125

#SPJ1

8 0
2 years ago
Rewrite with only sin x and cos x.
Annette [7]

Option A

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

<em><u>Solution:</u></em>

Given that we have to rewrite with only sin x and cos x

Given is cos 3x

cos 3x = cos(x + 2x)

We know that,

\cos (a+b)=\cos a \cos b-\sin a \sin b

Therefore,

\cos (x+2 x)=\cos x \cos 2 x-\sin x \sin 2 x  ---- eqn 1

We know that,

\sin 2 x=2 \sin x \cos x

\cos 2 x=\cos ^{2} x-\sin ^{2} x

Substituting these values in eqn 1

\cos (x+2 x)=\cos x\left(\cos ^{2} x-\sin ^{2} x\right)-\sin x(2 \sin x \cos x)  -------- eqn 2

We know that,

\cos ^{2} x-\sin ^{2} x=1-2 \sin ^{2} x

Applying this in above eqn 2, we get

\cos (x+2 x)=\cos x\left(1-2 \sin ^{2} x\right)-\sin x(2 \sin x \cos x)

\begin{aligned}&\cos (x+2 x)=\cos x-2 \sin ^{2} x \cos x-2 \sin ^{2} x \cos x\\\\&\cos (x+2 x)=\cos x-4 \sin ^{2} x \cos x\end{aligned}

\cos (x+2 x)=\cos x-4 \cos x \sin ^{2} x

Therefore,

\cos 3 x=\cos x-4 \cos x \sin ^{2} x

Option A is correct

7 0
3 years ago
If x=2 ,y=12 and x=4,y=3,find n and k if y is inversely proportional to x and k is a constant
Setler [38]

question:

x=2 ,<em>y</em>=12 and x=4,y=3,find n and k if y is inversely proportional to x and k is a constant

answer:

k=12 x=3 n=4

explain:

that the answer because the question said find the k and x and n so I find them

8 0
2 years ago
What is 73 divided by 8 equal in remainder???
olasank [31]

Answer:

9 remainder 1

Step-by-step explanation:

Not sure but uh... 73/8=9 remainder 1.

Is this what you are looking for?

4 0
2 years ago
Read 2 more answers
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