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boyakko [2]
3 years ago
10

Start at 125 create a pattern that subtract 6 from each number

Mathematics
1 answer:
Kay [80]3 years ago
5 0

If you want a sequence that subtracts 6 each time, then we'd have to have a nth term.

125, 119, 113 etc will be your pattern

Since we are subtracting 6 each time, the nth term will start with -6n

Then we see what that equation looks like: -6 as the first term, -12 as the second etc.

So we have to add a certain amount so the sequence starts at 125, which is to plus 131 to get to 125.

So now the pattern starts at 125 and decreases by 6 each term.

So U_n = -6n + 131

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What is 8+4(-6x+4)=-6-5x
sergeinik [125]
First do 4(-6x+4) distribute the 4 (So multiply each number in parenthesis by 4) so 8-24x+16=-6-5x

Then put like terms on each side

So do inverse of each number and do that to both sides to their like terms

8+16=-6+19x

Then bring the six over and add like terms

30 = 19x

Then divide 19 in each side and you get

X = 10/21
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answer is 2

Step-by-step explanation:

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icang [17]

Answer:

26°

Step-by-step explanation:

AC=tan 64

BC=x=sec 64

Using sine rule

(Sin 64)/tan 64 = (sin BAC)/sec 64

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Sin BAC= 1

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Find the sum of the even numbers shown below. 2+4+6+8+...+66+68+70​
Mandarinka [93]

The sum its +2

2+2=4

4+2=6

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5 0
3 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
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