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pantera1 [17]
2 years ago
13

The sum of 11 and t is 22

Mathematics
2 answers:
kumpel [21]2 years ago
8 0

Answer:

33

Step-by-step explanation:

LenKa [72]2 years ago
5 0

You will add the two numbers together and add the t as well to make 33t
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The 5th term in a geometric sequence is 40. The 7th term is 10. What is (are) the possible value(s) of the 4th term?
Lena [83]

Answer:

possible values of 4th term is 80 & - 80

Step-by-step explanation:

The general term of a geometric series is given by

a(n)=ar^{n-1}

Where a(n) is the nth term, r is the common ratio (a term divided by the term before it) and n is the number of term

  • Given, 5th term is 40, we can write:

ar^{5-1}=40\\ar^4=40

  • Given, 7th term is 10, we can write:

ar^{7-1}=10\\ar^6=10

We can solve for a in the first equation as:

ar^4=40\\a=\frac{40}{r^4}

<em>Now we can plug this into a of the 2nd equation:</em>

<em>ar^6=10\\(\frac{40}{r^4})r^6=10\\40r^2=10\\r^2=\frac{10}{40}\\r^2=\frac{1}{4}\\r=+-\sqrt{\frac{1}{4}} \\r=\frac{1}{2},-\frac{1}{2}</em>

<em />

<em>Let's solve for a:</em>

<em>a=\frac{40}{r^4}\\a=\frac{40}{(\frac{1}{2})^4}\\a=640</em>

<em />

Now, using the general formula of a term, we know that 4th term is:

4th term = ar^3

<u>Plugging in a = 640 and r = 1/2 and -1/2 respectively, we get 2 possible values of 4th term as:</u>

ar^3\\1.(640)(\frac{1}{2})^3=80\\2.(640)(-\frac{1}{2})^3=-80

possible values of 4th term is 80 & - 80

3 0
3 years ago
If ON = MN, find MN.<br> A 4<br> B. 5<br> C. 4 square root 2<br> D. 4 square root 30
elena-14-01-66 [18.8K]

Answer:

c

Step-by-step explanation:

ON=MN, so triangle onm is a right isosclese triangle, so ON=NM=4 square root 2

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3 years ago
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tresset_1 [31]

Answer:

17 5/8

Step-by-step explanation:

Thats the answer

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3 years ago
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A child has 4 wooden blocks. How many different ways can she stack 3 of them into a tower?
faltersainse [42]

Answer:

Step-by-step explanation:

To find the number of different ways she can stack 3 of them in a tower, we need to use the formula:

_{n}P_{k}=\dfrac{n!}{(n-k)!}

n = 4

k = 3

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_{n}P_{k}=\dfrac{4!}{1!}

_{n}P_{k}=\dfrac{4*3*2*1!}{(1)!}

_{n}P_{k}=\dfrac{4*3*2*1!}{1}

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Answer: I think its D

Explanation:I sort of made a coordinate graph

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