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2 years ago

These are the first six terms of a sequence starting with a1 = 80:

Mathematics

1 answer:

Nikitich [7]2 years ago

6 0

Answer:

$a _{n} = a( {r}^{n - 1} ) \\ a _{n} = 80(2 {}^{n - 1} ) \\ a _{n} = \frac{80. {2}^{n} }{ {2}^{ 1} } \\a _{n} = 40( {2}^{n} ) \\ a _{n} = ( {2}^{3} \times 5)( {2}^{n} ) \\ a _{n} = 5( {2}^{n + 3} )$

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Solving systems of equations by substitution -2x-3(6-11)=-7

Papessa [141]

Answer:

Step-by-step explanation:

Simplify both sides of the equation.

Subtract 15 from both sides.

Divide both sides by -2.

x = 11

3 0

3 years ago

Square A has a side length (2x-7) and Square B has side length (-4x+18). How much bigger is the perimeter of Square B than Squar

mihalych1998 [28]

How much bigger is B than A , so Side length B- A = (-4x+18) - (2x-7) = -6x+25
So perimeter = 4x length = 4x(-6x+25) = -24x+100 = 100-24x
How much bigger? 100-24X

__Brainliest if i helped!

4 0

3 years ago

Read 2 more answers

How much would it cost to buy 3 $13 battery packs and 3 $22 copper wires

Makovka662 [10]

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6 0

2 years ago

A jumbo crayon is composed of a cylinder with a conical tip. The cylinder is 12 cm tall with a radius of 1.5 cm, and the cone ha

s2008m [1.1K]

Answer:

Part 1) The lateral area of the cone is $LA=2\pi\ cm^{2}$

Part 2) The lateral surface area of the cylinder is $LA=36\pi\ cm^{2}$

Part 3) The surface area of the crayon is $SA=41.50\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the lateral area of the cone

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=1\ cm$

$l=2\ cm$

substitute

$LA=\pi (1)(2)$

$LA=2\pi\ cm^{2}$

Part 2) Find the lateral surface area of the cylinder

The lateral area of the cylinder is equal to

$LA=2\pi rh$

we have

$r=1.5\ cm$

$h=12\ cm$

substitute

$LA=2\pi (1.5)(12)$

$LA=36\pi\ cm^{2}$

Part 3) Find the surface area of the crayon

The surface area of the crayon is equal to the lateral area of the cone, plus the lateral area of the cylinder, plus the top area of the cylinder plus the bottom base of the crayon

Find the area of the bottom base of the crayon

$A=\pi[r2^{2}-r1^{2}]$