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2 years ago

Find the missing term in the following sequence ? 1, 4, 11, 26, 57, 120, ______

Mathematics

2 answers:

Advocard [28]2 years ago

7 0

Given sequence;

1, 4, 11, 26, 57, 120, _

The missing term =?

Solution;

We need to find the relationship between each term and then we can find the missing term;

1, 4, 11, 26, 57, 120, _ ;

let us start with the first two;

3 x 2 + 1 = 7 + 4 = 11

7 x 2 + 1 = 15 + 11 = 26

15 x 2 + 1 = 31 + 26 = 57

31 x 2 + 1 = 63 + 57 = 120

63 x 2 + 1 = 127 + 120 = 247

We see this pattern and can conclude that the missing term is 247

Marysya12 [62]2 years ago

5 0

Answer:

247

Step-by-step explanation:

1,4,11,26,57,120. see the pattern that emerges from the series:

4–1 = 3; 3–1 = 2 = 2^1

11–4 = 7; 7 - 3 = 4 = 2^2

26 - 11 = 15; 15 - 7 = 8 = 2^3

57 - 26 = 31; 31 - 15 = 16 = 2^4

120 - 57 = 63; 63 - 31 = 32 = 2^5

so the next number should be 64+63 = 127+120 = 247.

check: 247–120 = 127; 127–63 = 64 = 2^6. correct.

so the next number is 247. 2^n+(n-1)

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1 year ago

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almond37 [142]

Step 1

we know that

The equation of a circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +42x+38y-47=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+42x)+(y^{2}+38y)=47$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}$

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849$

Rewrite as perfect squares

$(x+21)^{2}+(y+19)^{2}=849$

The center of the circle is the point $(-21,-19)$

The radius of the circle is $\sqrt{849}\ units$

The answer Part a) is

The equation of the circle in standard form is equal to

$(x+21)^{2}+(y+19)^{2}=849$

The answer Part b) is

The center of the circle is the point $(-21,-19)$

The answer Part c) is

The radius of the circle is $\sqrt{849}\ units$

Let's verify each case to determine the solution of the second part of the problem

Step 2

we have

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Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+60x)+(y^{2}+14y)=-98$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}$

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851$

Rewrite as perfect squares

$(x+30)^{2}+(y+7)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 3

we have

$x^{2} +y^{2} +44x-44y+117=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+44x)+(y^{2}-44y)=-117$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}$

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Rewrite as perfect squares

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The radius of the circle is $\sqrt{851}\ units$

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Step 4

we have

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Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}$

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Rewrite as perfect squares

$(x-19)^{2}+(y+21)^{2}=728$

The radius of the circle is $\sqrt{728}\ units$

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Step 5

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Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square twice. Remember to balance the equation by adding the same constants to each side.

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1 year ago

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Serga [27]

Answer:

Yes

Step-by-step explanation:

ΔMNL ≅ ΔQNL by ASA or AAS

by ASA

Proof:

∠ LNM = ∠LNQ =90

LN = LN {Common}

∠MLN = ∠QLN {LN bisects ∠ L}

By AAS

∠Q + ∠QLN + ∠LNQ = 180 {Angle sum property of triangle}

∠Q + 32 + 90 = 180

∠Q + 122 = 180

∠Q = 180 -122 =

∠Q = 58

∠Q = ∠M

∠MNL =∠QNL = 90

LN = LN {common side}

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2 years ago

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