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earnstyle [38]
3 years ago
11

Please help I Will mark you Brainliest

Mathematics
2 answers:
alexdok [17]3 years ago
3 0
I believe the answer is D.
nalin [4]3 years ago
3 0
The answer is d I believe bb
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Given the 5th term is 103 and the come differences is -6 2/3 what is the 6th term?
LekaFEV [45]

Answer:

  102 1/3

Step-by-step explanation:

To find the 6th term, add the common difference to the 5th term:

  103 + (-6 2/3) = 102 1/3

6 0
4 years ago
Three numbers are in the ratio 3:9:10. If 10 is added to the last number, then the three numbers form an arithmetic progression.
MrRa [10]

Answer:

<em> The numbers are 6, 18, and 30 </em>

Step-by-step explanation:

If the three numbers are in the ratio of 3:9:10,

let the numbers be 3x, 9x and 10x.

<em>If 10 is added to the last number to form an arithmetic progression</em>

<em>Then, 3x 9x (10x+10) are the progression</em>

The common difference of an arithmetic progression (d) = T₂ - T₁ = T₃ - T₂

T₂-T₁ = T₃ - T₂ .............. Equation 1

Where T₁ = first term of the progression, T₂ = Second term of the progression, T₃ = third term of the progression

<em>Given: T₁ = 3x, T₂ = 9x, T₃ = 10x +10</em>

<em>Substituting these values into equation 1</em>

<em>9x-3x = (10x+10)-9x</em>

<em>Solving the equation above,</em>

<em>3x = 10+x</em>

<em>3x-x = 10</em>

<em>2x = 10</em>

<em>x = 10/2</em>

<em>x = 2.</em>

<em>Therefore the numbers are 6, 18, and 30 </em>

<em />

7 0
4 years ago
530 divided by 16 give the quotient and remainder
damaskus [11]
   
 
 
          33
      _____
16 | 530
       48
      -------
         50
         48
    ----------
            2

quotient = 33
remainder = 2








6 0
3 years ago
The figure below is the graph of the dimensions of a rectangle whose adjacent side lengths exhibit Inverse Variation (1,4) (2,2)
Stells [14]

Answer: true

Step-by-step explanation:

6 0
3 years ago
Prove that is here<br><img src="https://tex.z-dn.net/?f=1%20-%20cos%20%7B2%7Da%20%5Cdiv%201%20-%20sin%20a%7B2%7D%20%20%3D%20tan%
garri49 [273]

\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

<h3>LHS</h3>

\boxed{\sf \dfrac{cosA}{sinA}=cotA}

\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

\\ \sf\longmapsto 1-cot2A

\\ \sf\longmapsto 1-\dfrac{1}{tan2A}

\\ \sf\longmapsto \dfrac{tan2A-1}{tan2A}

\\ \sf\longmapsto tan2A

8 0
3 years ago
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