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mel-nik [20]
3 years ago
13

What is the fifth term of the geometric sequence a1=120, a2=36, a3=10.8, a6=0.2916?

Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0

The formula for geometric sequence is an = a1 r^(n-1) where r is the geometric factor and n is an interger. In the sequence given, r is equal to 3/10. In this case, an = 120* (3/10)^(n-1). Solving for a5, a5 = 120* (3/10)^(5-1) = 0.972
iragen [17]3 years ago
5 0
First we need to find the common ratio by dividing the second term by the first term. 36/120 = 3/10

an = a1 * r^(n-1)
n = term to find = 5
a1 = first term = 120
r = common ratio = 3/10
now we sub and solve
a5 = 120 * 3/10^(5 - 1)
a5 = 120 * 3/10^4
a5 = 120 * .0081
a5 = 0.972 <=== fifth term

or we could have just done this......since we multiply by 3/10 to find the next number...
a3 = 10.8......10.8 * 3/10 = 3.24 <== 4th term
3.24 * 3/10 = 0.972 <== fifth term
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Solve the following inequalities:<br><br> <img src="https://tex.z-dn.net/?f=-%28x%2B2%29%5E%7B2%7D%20%28x-3%29%28x%2B3%29%28x-4%
Salsk061 [2.6K]
<h2>Answer:</h2>

\boxed{(-\infty,-3] \ U \ [3,4] \ U \ \{-2\}}

<h2>Step-by-step explanation:</h2>

To solve this problem we need to use the Test Intervals for Polynomial. The following steps helps us to determine the intervals on which the values of a polynomial are entirely  negative or entirely positive.

1. Find all real zeros of the polynomial arranging them in increasing order from smallest to largest. We call these zeros the critical numbers of the polynomial.

Before we start applying this steps, let's multiply the entire inequality by -1 changing the direction of the inequality, so the result is:

(x+2)^2(x-3)(x+3)(x-4)\leq 0

So the real zeros are:

x_{1}=-3 \\ \\ x_{2}=-2 \\ \\ x_{3}=3 \\ \\ x_{4}=4

2. Use the real zeros of the polynomial to determine its test intervals.

(-\infty,-3) \\ \\ (-3,-2) \\ \\ (-2,3) \\ \\ (3,4)

3. Take one representative x-value in each test interval, then evaluate the polynomial at this value. If the value of the polynomial is negative, the polynomial will have negative values for every x-value in the interval. If the value of the polynomial is positive, the polynomial will have positive values for every x-value in the interval.

<u>Polynomial values</u>

___________________

(-\infty,-3): \ (-4+2)^2(-4-3)(-4+3)(-4-4)=-224 \ NEGATIVE \\ \\(-3,-2): \ (-2.5+2)^2(-2.5-3)(-2.5+3)(-2.5-4)=4.46 \ POSITIVE \\ \\ (-2,3): \ (-1+2)^2(-1-3)(-1+3)(-1-4)=40 \ POSITIVE \\ \\ (3,4): \ (3.5+2)^2(3.5-3)(3.5+3)(3.5-4)=-49.15 \ NEGATIVE \\ \\ (4,\infty): \ (5+2)^2(5-3)(5+3)(5-4)=784 \ POSITIVE

___________________

At x = -3, x = 3, and x = 4 we use brackets because the inequality includes the value at which the polynomial equals zero.

Finally, if you set x=-2:

(x+2)^2(x-3)(x+3)(x-4)\leq 0 \\ \\ (-2+2)^2(-2-3)(-2+3)(-2-4)\leq 0 \\ \\ 0\leq 0

So x=-2 is also a solution to the system. Finally, the solution is:

\boxed{(-\infty,-3] \ U \ [3,4] \ U \ \{-2\}}

8 0
3 years ago
Please help me with this question and please tell me how I would put this into a calculator please...
AnnyKZ [126]

Answer:

6

Step-by-step explanation:

\frac{2}{3}x+4=\frac{2}{3} *\frac{3}{1}+4 = 2+4=6

3 0
3 years ago
Solve for X. Opposite sides of the figure are parallel.
melisa1 [442]

Check the picture below.

4 0
3 years ago
10 y is directly proportional to the cube of x. Given that y = 1512 when x = 6, find the equation for the y in terms of x.​
sergiy2304 [10]

Answer:

{ \tt{y \:  \alpha  \:  {x}^{3} }} \\ { \tt{y = k {x}^{3} }} \\ { \bf{k \: is \: a \: proportionality \: constant}} \\ { \tt{1512 = (k \times  {6}^{3}) }} \\ { \tt{k = 7}} \\ { \boxed{ \bf{y = 7 {x}^{3} }}}

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3 years ago
Is a triangle with side lengths of 33 inches, 56 inches, and65 inches a right triangle? Explain your reasoning.
worty [1.4K]

Answer:

Yes

Step-by-step explanation:

Pythagorean's Theorum is that a^2 + b^2 = c^2

a and b are the sides of the triangle and c is the hypotenuse. 33x33+56x56=4225 sqrt4225 is 65. Everything checks out.

3 0
3 years ago
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