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3 years ago

9

Three bags of flour and two bags of sugar wait 32 pounds four bags of sugar wait 32 pounds all bags of flour wait to save it all

back to sugar wait to say what is the ratio of the weight of bag of flour two with bag of sugar

1 answer:

Contact [7]3 years ago

3 0

You worded this weird, but each bag of sugar is 8 pounds...

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3 years ago

Complete the statement to describe the expression a b c + d e f abc+defa, b, c, plus, d, e, f. the expression consists of terms,

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The statement of expressions is two terms and three factors.

<h3>The statement and expressions:</h3>

According to the question, the statement that teaches us about expression is made up of two terms, each of which has three factors.

The ABC and DEF expressions define the expression of two terms, each of which comprises two elements.

As a result, the answer is two terms and three factors.

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Variables are combined using multiplication signs to form the factor.
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1 year ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation $A=45e^{-0.05t}$ .

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate $A=15$ and solve for t as:

$15=45e^{-0.05t}$

$\frac{15}{45}=\frac{45e^{-0.05t}}{45}$

$\frac{1}{3}=e^{-0.05t}$

Now we will switch sides:

$e^{-0.05t}=\frac{1}{3}$

Let us take natural log on both sides of equation.

$\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})$

$-0.05t\cdot (1)=\text{ln}(\frac{1}{3})$

$-0.05t=\text{ln}(\frac{1}{3})$

$t=\frac{\text{ln}(\frac{1}{3})}{-0.05}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}$

$t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}$

$t=-\text{ln}(\frac{1}{3})\cdot 20$

$t=-(\text{ln}(1)-\text{ln}(3))\cdot 20$

$t=-(0-\text{ln}(3))\cdot 20$

$t=20\text{ln}(3)$

Therefore, it will take $20\text{ln}(3)$ years for area of the glacier to decrease to 15 square kilo-meters.

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2 years ago