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3 years ago

mark mowed 1/3 of a football field. he mowed equal amounts over 2 hours how much of a field did mark mow each hour?

Mathematics

1 answer:

grigory [225]3 years ago

4 0

Answer: 1/6

Step-by-step explanation: ok, so he had mowed 1/3 of a field in 2 hours. Take 1/3 and divide by 2 hours to get 1/6

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Find a third-degree polynomial equation with rational coefficients that has the roots 2 and -3 + i.

NeTakaya

Answer:

Option B

Step-by-step explanation:

Complex roots occur as conjugate pairs so the third root is -3 - i ( note that the sign changes from + to -).

So in factor form we have:-

(x - 2)(x - (-3 + i))(x - (-3 - i)) = 0 Let's expand the last 2 factors first:-

(x - (-3 + i))(x - (-3 - i))

= (x + 3 - i)(x + 3 + i)

= x^2 + 3x +ix + 3x + 9 + 3i - ix - 3i - i^2

= x^2 + 6x + 9 - (-1)

= x^2 + 6x + 10

Now multiplying by (x - 2):-

(x - 2)(x^2 + 6x + 10) = 0

x^3 + 6x^2 + 10x - 2x^2 - 12x - 20 = 0

x^3 + 4x^2 - 2x - 20 = 0 (answer)

Option B

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3 years ago

Is 0.0530 greater than 0.04?

SashulF [63]

Answer:

yes

Step-by-step explanation:

0.04 is equal to 0.0400

0.0530

0.0400

5 is higher than 4

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3 years ago

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How can you use a rate to compare the costs of two boxes of cereal that are different sizes?

Rainbow [258]

Answer:

✖ both sides

Step-by-step explanation:

you take both numbers and you multiply it

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Based on the pattern what are the next two terms for the pattern 4,12,20,28

alukav5142 [94]

Answer:

your next two numbers are 36 and 44

Step-by-step explanation:

because you are adding on 8 to each number

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3 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

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

where

r2 is the radius of the cylinder

r1 is the radius of the cone

substitute

$A=\pi[1.5^{2}-1^{2}]$

$A=1.25\pi\ cm^{2}$

Find the area of the top base of the cylinder

$A=\pi(1.5)^{2}=2.25\pi\ cm^{2}$

Find the surface area

$SA=2\pi+36\pi+2.25\pi+1.25\pi=41.50\pi\ cm^{2}$

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