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Gnom [1K]
3 years ago
14

Identify the 34th term of the arithmetic sequence 2, 7, 12

Mathematics
1 answer:
torisob [31]3 years ago
5 0
The equation is AN=A1+D(N-1) A1 being the first number (2) D being the difference (in this case 5) and N being the nth term to find (34) so here you have AN=2+5(34-1) then AN=2+5(33). the answer here is 167.
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bogdanovich [222]
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4 0
3 years ago
Read 2 more answers
Write down the answers to a,b,c,d
aniked [119]

Answer:

(A) 1

(B) -2

(C) 3.5

(D) -0.5

Step-by-step explanation:

We can treat each thermometer like a vertical number line and read the values on each.

A is right on 1.

B is right on -2.

C is in the middle of 3 and 4, so 3.5

D is in the middle of 0 and -1, so -0.5

Hope this helped!

3 0
3 years ago
Find the nth term or this sequence<br><br> 8 12 16 20 24 ...
scoray [572]
8\\8+4\\8+4+4\\8+4+4+4\\...\\8+4(n-1)

8+4(n-1)
8 0
3 years ago
Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

B^2 = 100 + 144 - 240*0.86602540378

B^2 = 100 + 144 - 207.846096907

B^2 = 36.153903093

Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

-80 = - 144 *Cos\alpha

Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

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3 years ago
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Step-by-step explanation:

7 0
3 years ago
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