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IRISSAK [1]
3 years ago
9

how to use the explicit formula an=a1+(n-1)d to find the 500th term of the sequence 24,30,36,42,48,...

Mathematics
2 answers:
Leni [432]3 years ago
7 0
A1=24 
<span>n=500(seq #) </span>
<span>d=how much u add each time=6 </span>

<span>a(n)=24+(500-1)(6)=3018</span>
Fantom [35]3 years ago
4 0

Answer:

3517

Step-by-step explanation:

Apex quiz, I got it right

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Use a trigonometric ratio to find the value of x. Round your answer to the nearest<br> tenth.
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Step-by-step explanation:

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3 years ago
Determine the equation of the circle graphed below. <br><br> ( please help me )
Brums [2.3K]

Answer:

Equation : \\(x+3)^2 + ( y -2)^2 = 36

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: <em><u>Find the centre</u></em>

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

                        Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) =  (-3, 2)

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : <u>Find radius</u>

<u></u>Radius = \frac{Diameter }{2}<u></u>

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

 Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\

                 = \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12

Therefore ,

       Radius = \frac{12}{2} = 6

Step 3 : <u>Equation of the circle</u>

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

       (x - h)^2+(y -k)^2 = r^2

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

    (x - (-3))^2 +  (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36

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2 years ago
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8 0
3 years ago
Read 2 more answers
Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ
solong [7]

Answer:

The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.    

Step-by-step explanation:

We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line  x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.    

3 0
3 years ago
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