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3 years ago

Describe the following sequence as arithmetic, geometric or neither. 2, 4, 6, 12, 14. . . .

Mathematics

2 answers:

Pepsi [2]3 years ago

7 0

It Is neither arithmetic or geometric

murzikaleks [220]3 years ago

5 0

If it was arithmetic each term would be the same difference from the previous term...

4-2 does not equal 12-6 so this is not an arithmetic sequence.

If it was a geometric sequence each term would be a constant ratio when compared to the previous term...

4/2 does not equal 6/4 so this is not a geometric sequence.

So this sequence is neither.

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3 years ago

.. Which of the following are the coordinates of the vertices of the following square with sides of length a?

atroni [7]

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Step-by-step explanation:

Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

To find the sides of a square, let us use the distance formula,

$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length OS }=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } S T=\sqrt{(a-a)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } T W=\sqrt{(0-a)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a\\&\text { Length } O W=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

Thus, the square with vertices $O(0,0), S(a,0), T(a,a), W(0,a)$ has sides of length a.

Thus, the correct answers are option a and option d.

8 0

3 years ago

Can plz help answer number 4

enot [183]

Answer:

It was 6 7/8 pounds of fruit in the fruit salad.

Step-by-step explanation:

Regina bought in total (in pounds) of fruit: B

$B=3\frac{1}{2}+4\frac{3}{4}\\ B=\frac{3(2)+1}{2}+\frac{4(4)+3}{4}\\ B=\frac{6+1}{2}+\frac{16+3}{4}\\ B=\frac{7}{2}+\frac{19}{4}\\ B=\frac{7}{2}.1+\frac{19}{4}\\ B=\frac{7}{2}.\frac{2}{2}+\frac{19}{4}\\ B=\frac{7(2)}{2(2)}+\frac{19}{4}\\ B=\frac{14}{4}+\frac{19}{4}\\ B=\frac{14+19}{4}\\ B=\frac{33}{4}$

Regina bought in total 33/4 pounds of fruit

She used all (33/4 pounds) but 1 3/8 pounds of the fruit in a fruit salad, then she used in the fruit salad (in pounds): S

$S=B-1\frac{3}{8}\\ S=\frac{33}{4}-\frac{1(8)+3}{8}\\ S=\frac{33}{4}-\frac{8+3}{8}\\ S=\frac{33}{4}-\frac{11}{8}\\ S=\frac{33}{4}.1-\frac{11}{8}\\ S=\frac{33}{4}.\frac{2}{2}-\frac{11}{8}\\ S=\frac{33(2)}{4(2)}-\frac{11}{8}\\ S=\frac{66}{8}-\frac{11}{8}\\ S=\frac{66-11}{8}\\ S=\frac{55}{8}\\ S=\frac{48+7}{8}\\ S=\frac{48}{8}+\frac{7}{8}\\ S=6+\frac{7}{8}\\ S=6\frac{7}{8}$

It was 6 7/8 pounds of fruit in the fruit salad.

7 0

3 years ago