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3 years ago

Simplify this question

Mathematics

2 answers:

Ne4ueva [31]3 years ago

8 0

$4 + \sqrt{ - 16} \\ = 4 + \sqrt{ - 2 \times 2 \times 2 \times 2} \\ = 4 + 2\sqrt{ - 4} \\ = 2(2 + \sqrt{ - 4} )$

Hope you could get an idea from here.

Doubt clarification - use comment section.

lutik1710 [3]3 years ago

6 0

$\\ \sf\longmapsto 4+√-16$

$\\ \sf\longmapsto 4+√16i$

$\\ \sf\longmapsto 4+4i$

Take 4common

$\\ \sf\longmapsto 4(1+i)$

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Please help me:) 30 points

Brut [27]

Answer:

C(N(h)) = 1400h +530

Step-by-step explanation:

x = N(h), so substitute in the expression for N(h) to get x

C(N(h)) = 35(40h) + 530

=> 1400h + 530

now pray for my lazy brain to check work it is correct

5 0

3 years ago

Read 2 more answers

Help? y - intercepts <br> What is the y-intercept of the equation 2x + 3y = 12?

Nookie1986 [14]

Answer:

2x + 3y = 12 Is the standard form for linear equations. To determine the slope, solve the equation for y in order to convert to the slope - intercept from7 y = mx + b where is m the slop and b is the y - intercept.

2x + 3y = 12

subtract 2x from both sides.

3y = -2x + 12

Divide both sides by 3.

y= - 2/3 x + 12/3 =

y= - 2/3x + 4

The slope is - 2/3

5 0

3 years ago

Please help me with this question

Setler [38]

Step-by-step explanation:

${ ({y}^{2} )}^{4} . \: {( {y}^{3}) }^{2}$

Since,

${ ({a}^{x} )}^{y} = {a}^{xy}$

Hence,

${y}^{2 \times 4} . {y}^{3 \times 2}$

$= {y}^{8} . {y}^{6}$

Since,

${a}^{x} . {a}^{y} = {a}^{x + y}$

Hence,

$= {y}^{8 + 6}$

$= {y}^{14} (ans)$

6 0

3 years ago

our bedroom ceiling is 10 feet high and is 2/3 as high as the living room ceiling. Write and solve an equation to find the heigh

Doss [256]

I know this probably isn't helpful to you, but the answer is 15. I suck at showing my work. So I really can't help you there. Sorry I couldn't be more help.

8 0

3 years ago

6n + n^7 is divisible by 7 and prove it in mathematical induction<br>

kompoz [17]

Answer:

Apply induction on $n$ (for integers $n \ge 1$ ) after showing that $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, \dots,\, 6 \rbrace$ .

Step-by-step explanation:

Lemma: $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ .

Proof: assume that for some $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ , $\genfrac{(}{)}{0}{}{7}{j}$ is not divisible by $7$ .

The combination $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is known to be an integer. Rewrite the factorial $7!$ to obtain:

$\displaystyle \begin{pmatrix}7 \\ j\end{pmatrix} = \frac{7!}{j! \, (7 - j)!} = \frac{7 \times 6!}{j!\, (7 - j)!}$ .

Note that $7$ (a prime number) is in the numerator of this expression for $\genfrac{(}{)}{0}{}{7}{j}\!$ . Since all terms in this fraction are integers, the only way for $\genfrac{(}{)}{0}{}{7}{j}$ to be non-divisible by $7\!$ is for the denominator $j! \, (7 - j)!$ of this expression to be an integer multiple of $7\!\!$ .

However, since $1 \le j \le 6$ , the prime number $\!7$ would not a factor of $j!$ . Similarly, since $1 \le 7 - j \le 6$ , the prime number $7\!$ would not be a factor of $(7 - j)!$ , either. Thus, $j! \, (7 - j)!$ would not be an integer multiple of the prime number $7$ . Contradiction.

Proof of the original statement:

Base case: $n = 1$ . Indeed $6 \times 1 + 1^{7} = 7$ is divisible by $7$ .

Induction step: assume that for some integer $n \ge 1$ , $(6\, n + n^{7})$ is divisible by $7$ . Need to show that $(6\, (n + 1) + (n + 1)^{7})$ is also divisible by $7\!$ .

Fact (derived from the binomial theorem $(\ast)$ ):

$\begin{aligned} & (n + 1)^{7} \\ &= \sum\limits_{j = 0}^{7} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] && (\ast)\\ &= \genfrac{(}{)}{0}{}{7}{0} \, n^{0} + \genfrac{(}{)}{0}{}{7}{7} \, n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] \\ &= 1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\end{aligned}$ .

Rewrite $(6\, (n + 1) + (n + 1)^{7})$ using this fact:

$\begin{aligned} & 6\, (n + 1) + (n + 1)^{7} \\ =\; & 6\, (n + 1) + \left(1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \\ =\; & 6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \end{aligned}$ .

For this particular $n$ , $(6\, n + n^{7})$ is divisible by $7$ by the induction hypothesis.

$\sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]$ is also divisible by $7$ since $n$ is an integer and (by lemma) each of the coefficients $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7\!$ .

Therefore, $6\, (n + 1) + (n + 1)^{7}$ , which is equal to $6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right)$ , is divisible by $7$ .

In other words, for any integer $n \ge 1$ , if $(6\, n + n^{7})$ is divisible by $7$ , then $6\, (n + 1) + (n + 1)^{7}$ would also be divisible by $7\!$ .

Therefore, $(6\, n + n^{7})$ is divisible by $7$ for all integers $n \ge 1$ .

6 0

2 years ago