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enyata [817]
3 years ago
15

the value of the nth term in a sequence of numbers is given by the expression 2n-3. What is the 8th term in the sequence

Mathematics
2 answers:
Debora [2.8K]3 years ago
8 0

Answer:

2n-3=0

2n=3

n=3/2

therefore answer is 3/2

ValentinkaMS [17]3 years ago
8 0

Answer:

bro I didn't even look up this question and it showed up so  I don't know dude

Step-by-step explanation:

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The side of a rhombus is 15 centimeters long, and the length of its longer diagonal is 24.6 centimeters. Find the area.
Mekhanik [1.2K]
<span>A = 175.03.
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The area of a rhombus is found by multiplying 1/2 <span>by the </span>diagonal<span> by the </span>other diagonal<span>.</span>
4 0
3 years ago
a pizza costs $11 and $1.50 per topping write a equation for cost in dollars and the nunber of toppings ​
romanna [79]

Answer:

y= 11 + 1.50x. (y is the price, x is number of toppings)

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Find the common ratio for this geometric sequence 8, 2, 1/2, 1/8, 1/32
kompoz [17]

Answer:

Multiply 1/4 ; dividing 4

Step-by-step explanation:

Note that in each number going from left to right, it decreases by dividing 4 each time (or multiplying 1/4)

8/4 = 2 ; 2/4 = 1/2 ; (1/2)/(4) = 1/8 ; etc.

(8)(1/4) = 2 ; (2)(1/4) = 1/2 ; etc.

~

7 0
3 years ago
Read 2 more answers
If X2-26X+69=0 and if X-5&gt;0 what is the value of X
denpristay [2]

Answer:

X=+260

Step-by-step explanation:

In the picture. I hope that it's a clear.

4 0
3 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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