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3 years ago

Find (1/3 + 5/6 - 1/6) / (4/5)

Mathematics

2 answers:

Marta_Voda [28]3 years ago

5 0

Answer:

$\frac{5}{4}$

Step-by-step explanation:

$(\frac{1}{3} + \frac{5}{6} - \frac{1}{6}) \div \frac{4}{5}\\\\( \frac{2 + 5 - 1}{6}) \div \frac{4}{5} \\\\(\frac{6}{6}) \div \frac{4}{5}\\\\\frac{1}{\frac{4}{5}}\\\\\frac{5}{4}$

Nataliya [291]3 years ago

3 0

Answer:

5/4

Step-by-step explanation:

A simpler approach:

(1/3+5/6-1/6):(4/5)=

(1/3+4/6):(4/5)=

(1/3+2/3):(4/5)=

1:(4/5)=

5/4

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PLSSSSSSSSSSSSSSSSSSSSSSSS HELP ME!!

zzz [600]

Answer:

For the top table:

[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |

[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |

For the bottom table:

k = 5

[x] | 3.1 | 2.5 | 1.2 | 0.9 | 0.14 | 0.06 | 0.02 |

[y] | 15.5 | 12.5 | 6 | 4.5 | 0.7 | 0.3 | 0.1 |

3 0

2 years ago

Don works in home construction. He determined that the function t(A)=144A/9.252 gives the total number of 9 in. tiles needed to

Oksanka [162]

Answer:

6,226 tiles of 9 in are needed to cover an area of 400 square feet

Step-by-step explanation:

Let

A -----> an area in square feet

t(A) ---> the total number of 9 in. tiles needed

we have

$t(A)=\frac{144A}{9.252}$

For A=400 ft^2

substitute in the equation and solve for t

$t(400)=\frac{144(400)}{9.252}$

$t(400)=6,226\ tiles$

That means -----> 6,226 tiles of 9 in are needed to cover an area of 400 square feet

6 0

2 years ago

A closed box has a square base with side length feet and height feet given that the volume of the box is 34 ft.³ express the sur

Deffense [45]

Answer:

$\dfrac{2s^3+136}{s}$

Step-by-step explanation:

Let the side length of the square base =s feet

Let the height of the box = h

Given that the volume of the box = $34$ ft^3$

Volume of the box = $s^2h$

Then:

$s^2h=34$ ft^3\\$Divide both sides by s^2\\h=\dfrac{34}{s^2}$

Surface Area of a Rectangular Prism =2(lb+bh+lh)

Since we have a square base, l=b=s feet

Therefore:

Surface Area of our closed box $= 2(s^2+sh+sh)$

$S$urface Area= 2s^2+4sh\\Recall: h=\dfrac{34}{s^2}\\$Surface Area= 2s^2+4s\left(\dfrac{34}{s^2}\right)\\=2s^2+\dfrac{136}{s}\\$Surface Area in terms of length only=\dfrac{2s^3+136}{s}$