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Sergio039 [100]
2 years ago
13

The difference of three times a number and five is four. Find the number

Mathematics
1 answer:
givi [52]2 years ago
4 0

The difference of three times a number and five is four. Then the number is 3

<u>Solution:</u>

Given that,

Difference of three times a number and five is four

We have to find the number

Let us frame a equation using the given information and solve the sum

Let the number be "n"

Difference of three times a number and five = 4

Three times n - 5 = 4

3n - 5 = 4

3n = 9

On dividing 9 by 3 we get 3

n = 3

Thus the required number is 3

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In-s [12.5K]

Answer:

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

Step-by-step explanation:

Given that:

\iiint_W (x^2+y^2) \ dx \ dy \ dz

where;

the top vertex = (0,0,1) and the  base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \  dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

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6 0
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5 - 3(-2) + 1-3

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5 0
2 years ago
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For each question (a–e) below, choose one of the answers (i–viii); explain your choice.
Natasha2012 [34]

Answer:

Step-by-step explanation:

given that a deck of cards is shuffled.

we know in a deck there are 52 cards, 13 cards of each variety spade, clubs hearts and dice.  Red are 26 and black are 26.  kings, will be 4.

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Answer with Step-by-step explanation:

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