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Ket [755]
3 years ago
10

{7,21,63,189,..} write a rule for the geometric sequence

Mathematics
1 answer:
sweet [91]3 years ago
8 0

Answer:

a7=a3 ^n-1

Step-by-step explanation:

:) Good day

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Determine if the triangles are similar. If similar, state how and complete the similarity stated.​
Alenkinab [10]

Answer:

a record of all the mail that a mailbox identifies as Junk

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3 years ago
Find the composite area of the shaded region
Anvisha [2.4K]

Answer:

what region

Step-by-step explanation:

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3 years ago
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The volume of a pyramid varies jointly as the height and the area of the base. If a pyramid has the measurements V=1144
Naddika [18.5K]

Answer:

  • 11040 m³
  • k ≈ 0.33
  • V = (1/3)Bh

Step-by-step explanation:

The given relation is ...

  V = kBh . . . . . for some base area B, height h, and constant of variation k

We are given length and width of the base so we presume it is a rectangle.

  B = l·w = 8·11 = 88 . . . . square meters

The given volume tells us the value of k:

  1144 = k(88)(39) . . . . . . cubic meters

  1144/3432 = k = 1/3 ≈ 0.33

The value of k is about 0.33.

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Then the volume of the larger pyramid is ...

  V = (1/3)(15 m)(46 m)(48 m) = 11,040 m³

The general relationship is ...

  V = 1/3Bh

7 0
3 years ago
Is -3y = 5x + 4 perpendicular
aivan3 [116]

Answer:

no it is not perpedicular

Step-by-step explanation:

I plugged the equation into a graphing calculator and the line does not appear to be perpendicular

5 0
3 years ago
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Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
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