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4 years ago

If you can buy 1/3 of a box of chocolates for 6 dollars, how much can you purchase for 4 dollars

Mathematics

1 answer:

zysi [14]4 years ago

7 0

Answer:

2/9

Step-by-step explanation:

Since 1/3 of a box of chocolates costs 6 dollars, you can buy (4×(1/3))/6=2/9 of a box of chocolates!

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Find the area of the shaded region in square units. Show your reasoning.

motikmotik

Answer:

40 square units

Step-by-step explanation:

First of all, lets say that square has side $l$ , so, the area unit is $l^2$

the diagonal's square is $l\sqrt{2}$

CALCULATION OF TRIANGLES'S AREA (there are 4 triangles)

$A_{triangles}=4*base*heigh*0.5=2*(l\sqrt{2} )(2l\sqrt{2} )=8l^2$

CALCULATION OF MAIN SQUARE AREA

$A_{square}=side*side=(4\sqrt{2} l)(4\sqrt{2} l)=32l^2$

TOTAL AREA

$A_{total}=A_{triangles}+A_{square}=8l^2+32l^2=40l^2$

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3 years ago

Would it be<br> A.<br> B.<br> C.<br> D.

gtnhenbr [62]

Answer:

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2 years ago

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How to solve the math equation

ddd [48]

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3 years ago

Express 28:7 in the form of n:1

mixas84 [53]

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4 years ago

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago