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Virty [35]
3 years ago
14

What is the measure of angle NQP

Mathematics
2 answers:
vichka [17]3 years ago
5 0
134+2x+(2x+2)=180(angle sum of triangle)
136+4x=180
4x=44
x=11°
so angle NQP=2(11)+2
=24°
olga2289 [7]3 years ago
4 0
What grade level are you in to need to know this
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3 questions pls help
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4 0
3 years ago
Help and you get 20 points and you will feel good i hope
Sergeu [11.5K]

Answer:

Change your greater than to less than

Step-by-step explanation:

The -1 1/4 is farther away from 0 than -1, which means it's less than -1.

3 0
3 years ago
What is 2 9/10 as a decimal.
ziro4ka [17]
2.9 or 2.90
2 is a whole number
9/10 is a fraction so you divide. so 9 because it is like 90/100. Basically it is a decimal and then a 9 :)
3 0
3 years ago
Read 2 more answers
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
The dean of admissions in a large university has determined that the scores of the freshman class in a mathematics test are norm
fomenos

Answer:

1

Step-by-step explanation:

We know that the mathematics test scores are normally distributed with mean μ=82 and standard deviation σ=8.

We know that the central limit theorem states that a selected sample from a normal distribution with mean μ and standard deviation σ is also normally distributed with mean μxbar and standard deviation σxbar.

Also , the sample mean of sampling distribution is μxbar=μ, where μ is population mean.

The standard deviation of sampling distribution is σxbar=σ/√n where n is the sample size and σ is population standard deviation.

The given sample size=n=64.

So, the required standard deviation of sampling distribution is

σxbar=σ/√n

σxbar=8/√64

σxbar=8/8

σxbar=1.

5 0
3 years ago
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