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3 years ago

Find the 9th term in the following arithmetic sequence -7, -1, 5, 11

Mathematics

2 answers:

Masja [62]3 years ago

6 0

Answer:

Step-by-step explanation:

calculate the difference between each number and find out how many are left in the sequence and multiply by that

in this case the difference was 6 and you needed 5 more until you hit the ninth number so 5*6+30 30+11=41

inn [45]3 years ago

3 0

Answer:

Step-by-step explanation:

1. Multiply 5 by 6 in order to get to the 9th number. (5 places away, 4 were given.) Also, notice that the difference between all of the numbers are consistently 6.

2. Add 30 to 11, which gives you 41.

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CLEAR HIGHLIGHTING

Tasya [4]

Yikes I think 3 or 6

7 0

3 years ago

Complete the expressions to find the perimeter PP of a rectangle that is 7 inches long by 5 inches wide.

neonofarm [45]

PP = 2L + 2W. Here, PP = 2(7 inches) + 2(5 inches) = 24 inches.

7 0

3 years ago

Not quite sure how to work this out

Zina [86]

you have the hypotenuse and the opposite side so you will use sin
sin58= (30/q)
multiply both sides by q
q(sin58)=30
divide by sin58 in your calculatior
q=(30/sin58)
^ I don't have a calculator with me so you will need to plug that in

5 0

3 years ago

Will give BRAINLEST :)

DiKsa [7]

Answer:

Step-by-step explanation:

This explanation mostly depends on what you're learning right now. The first way would be to convert this matrix to a system of equations like this.

g + t + k = 90

g + 2t - k = 55

-g - t + 3k = 30

Then you solve using normal methods of substitution or elimination. It seems to me that elimination is the quickest method.

g + t + k = 90

-g - t + 3k = 30

____________

0 + 0 + 4k = 120

4k = 120

k = 30

No you can plug this into the first two equations

g + t + (30) = 90

g + t = 60

and

g + 2t - (30) = 55

g + 2t = 85

now use elimination again by multiplying the first equation by -1

g + 2t = 85

-g - t = -60

_________

0 + t = 25

t = 25

Now plug those both back into one of the equations. I'll just do the first one.

g + (25) + (30) = 90

g = 35

Therefore, we know that Ted spent the least amount of time on the computer.

The second method is using matrix reduction and getting the matrix in the row echelon form, therefore solving using the gauss jordan method. If you would like me to go through this instead, please leave a comment.

3 0

3 years ago

I don't know if this is right... please someone help mee

worty [1.4K]

For the first circle, let's use the pythagorean theorem

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17$

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.

now, let's check for second circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938$

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.

let's check the third circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}$

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.

6 0

3 years ago