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Alexeev081 [22]
3 years ago
11

What's a consecutive side

Mathematics
1 answer:
vodomira [7]3 years ago
6 0
Consecutive sides are any two sides that meet at an endpoint or an angle. Recognizing consecutive sides is important for many functions in geometry, as it helps identify segments and angles.
You might be interested in
The exterior angle of a regular polygon is 15 degrees how many sides does it have
Komok [63]
The formula to use is
n = 360/E
where 
n = number of sides
E = measure of exterior angle

In our case, E = 15 so 
n = 360/E
n = 360/15
n = 24

This regular polygon has 24 sides
6 0
3 years ago
Solve x^2+20x+100=50
uysha [10]
1. You con solve the quadratic equation x^2+20x+100=50<span> by following the proccedure below:
 
 2. Pass the number 50 from the right member to the left member. Then you obtain:
 
 x^2+20x+100-50=0
</span><span> x^2+20x+50=0
</span><span> 
 3. Then, you must apply the quadratic equation, which is:
 
 x=(-b±√(b^2-4ac))/2a
 
 </span><span>x^2+20x+50=0
</span><span> 
 a=1
 b=20
 c=50
 

  4. Therefore, when you substitute the values into the quadratic equation and simplify ir, you obtain that the result is:
 
  -10</span>±5√2   (It is the last option).
6 0
3 years ago
Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
A one-tailed test is a a. hypothesis test in which rejection region is in one tail of the sampling distribution b. hypothesis te
garri49 [273]

Answer:

Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution

Step-by-step explanation:

One Tailed Test:

  • A one tailed test is a test that have hypothesis of the form

H_0: \bar{x} = \mu\\H_A: \bar{x} < \mu\text{ or } \bar{x} > \mu

  • A one-tailed test is a hypothesis test that help us to test whether the sample mean would be higher or lower than the population mean.
  • Rejection region is the area for which the null hypothesis is rejected.
  • If we perform right tailed hypothesis that is the upper tail hypothesis then the rejection region lies in the right tail after the critical value.
  • If we perform left tailed hypothesis that is the lower tail hypothesis then the rejection region lies in the left tail after the critical value.

Thus, for one tailed test,

Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution

6 0
3 years ago
the number is 57,733 contains two sets of digits in which one digit is ten times as great as the other. what are the values of t
Afina-wow [57]
In 57,733 the values are:

5 is for tens
The first 7 is for ones
The second 7 is for tenths
The first 3 is for hundredths
The second 3 is for thousandths

Two set of digit in which one digit is ten times as great as the other are 3 and 3, as well as 7 and 7

It's because 0,03 is ten times as much as 0,003 and 7 is ten times as much as 0,7.
8 0
3 years ago
Read 2 more answers
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