What is the value of Fraction 1 over 2x3 + 3.4y when x = 4 and y = 2?

Mathematics

1 answer:

forsale [732]4 years ago

4 0

Plug in x and y

1 / [2(4)3+3.4(2)]
1 / 24+6.8
1 / 30.8

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Find the 8th term of the geometric sequence whose common ratio is 2/3 and whose first term is 7

motikmotik

Answer:

The 8th term of the sequence is 896/2187.

Step-by-step explanation:

We want to find the 8th term of a geometric sequence whose common ratio is 2/3 and whose first term is 7.

We can write a direct formula. Recall that the direct formula of a geometric sequence is given by:

$\displaystyle x_{n} = a\left(r\right)^{n-1}$

Where a is the initial term and r is the common ratio.

Substitute:

$\displaystyle x_{n} = 7\left(\frac{2}{3}\right)^{n-1}$

To find the 8th term, let n = 8. Substitute and evaluate:

$\displaystyle \begin{aligned} x_{8} &= 7\left(\frac{2}{3}\right)^{(8) - 1} \\ \\ &= 7\left(\frac{2}{3}\right)^{7} \\ \\ &= 7\left(\frac{128}{2187}\right) \\ \\ &= \frac{896}{2187} = 0.4096...\end{aligned}$

In conclusion, the 8th term of the sequence is 896/2187.

8 0

3 years ago

Classify the triangle shown below. check all that apply? 60 60 60

Stells [14]

Your Answer Is....

A) Acute, C) Equilateral, And D) Isosceles.

Why?

Acute = All Sides Less Than 90°. Equilateral = All Sides The Same. Isosceles = 2 Or 3 Sides The Same. Therefore, A, C, And D Are Your Answers.

≈≈≈≈ Glad I Could Help, And Good Luck! ≈≈≈≈

My Name: AnonymousGiantsFan

6 0

3 years ago

Read 2 more answers

Suppose you have 18 objects (10 of type A, 5 of type B, and 3 of type C). Objects of type A are indistinguishable from each othe

Paladinen [302]

Answer:

$\binom{18}{5}= 8568$

Step-by-step explanation:

Note that we have in total 18 items. Even though we are given information regarding the amounts of items per type, the general question asks the total number of ways in which you can pick 5 out of the 18 objects, without any restriction on the type of chosen items. Therefore, the information regarding the type is unnecessary to solve the problem.

Recall that given n elements, the different ways of choosing k elements out of n is given by the binomial coefficient $\binom{n}{k})$ .