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4 years ago

The points shown on the graph represent the numbers in a geometric sequence.What is the initial value of the sequence? 1 2 3 8

Mathematics

2 answers:

Irina-Kira [14]4 years ago

5 0

Answer:

Step-by-step explanation:

It starts with number one, so one must be the initial value of the sequence.

professor190 [17]4 years ago

3 0

Step-by-step explanation:

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the length of a rectangle is 14 meters. the perimeter of the rectangle is 44 meters. what is the width?

fomenos

Answer:

Step-by-step explanation:

6 0

2 years ago

Which expression is equivalent to (m8n-4) 1/4

ser-zykov [4K]

Answer:

D. $\dfrac{m^2}{n}$

Step-by-step explanation:

I assume this is your problem:

$(m^8n^{-4})^{\frac{1}{4}}$

If so, then this is the solution.

$(m^8n^{-4})^{\frac{1}{4}} =$

$= m^{8 \times \frac{1}{4}} n^{-4 \times \frac{1}{4}}$

$= m^2 n^{-1}$

$= \dfrac{m^2}{n}$

4 0

3 years ago

Read 2 more answers

Which situation below correctly represents the algebraic expression 4 + 5g? A. Nikki has 4 sticks of gum and Breanne has 5 times

zhenek [66]

Answer:

Step-by-step explanation:

The 4 is the base price. Paying 5g or $5 per game is the add on

7 0

3 years ago

the world's tallest mountain, if measured from base to peak, is Mauna Kea in Hawaii. The base has an elevation of -19680 feet. T

saveliy_v [14]

Answer:

33476 feet tall

Step-by-step explanation:

If we add the elivation of the base and the peak together we get 33476

3 0

3 years ago

Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.(Hint: since you are told

ANTONII [103]

Answer:

See proof below

Step-by-step explanation:

We will use the hint. The statement of the hint holds true, as the linear span of a set of vectors T is equal to the set of linear combinations of vectors in T.

Denote the linear span of vectors with the curly brackets < >, that is, $span\{v_1,v_2,v_3\}:=$

Let $u\in$ , then u is a linear combination of v1,v2,v3, that is, there exist scalars $c_1,c_2,c_3\in \mathbb{R}$ such that $u=c_1v_1+c_2v_2+c_3v_3$ . Multiply by 3 in both sides to get $3u=3c_1v_1+3c_2v_2+3c_3v_3=d_1v_1+d_2v_2+d_3v_3$ , with $d_i=3c_i,i=1,2,3$

Since $c_1,c_2,c_3\in \mathbb{R}$ , $d_1,d_2,d_3\in \mathbb{R}$ as real numbers are closed under multiplication. Therefore 3u is a linear combination of the vectors $v_1,v_2,v_3$ , that is, $3u\in$

8 0

3 years ago