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Korvikt [17]
3 years ago
15

Eric has a trolley. He wants to use that trolley to transport wooden blocks. The size of trolley is 3,960,000 square centimeter.

He found the wooden blocks to have a density of 0.0029 blocks per square centimeter. How many total wooden blocks can he carry in the trolley?
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
6 0

Answer:

1,365,517,241 blocks

Step-by-step explanation:

number of wooden blocks that can be carried = 3960000/0.0029=1,365,517,241 blocks

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Answer:

<em>Answer: 1/4</em>

Step-by-step explanation:

<u>Algebraic Operations</u>

The width of a rectangle is given as W=5x+3 and the length of the rectangle is given as L=20x+12.

It's required to determine the ratio of the width to the length, that is:

\displaystyle   \frac{W}{L}=\frac{ 5x+3}{20x+12}

This fraction can be simplified if we factor the expression in the denominator. The LCF of 20x and 12 is 4, thus:

\displaystyle   \frac{W}{L}=\frac{ 5x+3}{4(5x+3)}

Now note we have the exact same expression up and down, thus simplifying:

\displaystyle   \frac{W}{L}= \frac{1}{4}

Answer: 1/4

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3 years ago
What is the answer to 4x + 27
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Answer:

Step-by-step explanation:

4x+27

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4       4

x=6.75 or x=7

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2 years ago
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Answer:

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3 years ago
How is this solved using trig identities (sum/difference)?
GenaCL600 [577]
FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

cos \beta = -\frac{1}{2}  \sqrt{3}

tan \beta= \frac{1}{3}  \sqrt{3}

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2}  \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )
sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )
sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )
cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )
cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )

Find tan (α - β)
tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha  tan \beta }
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}

Simplify the denominator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5\sqrt{3}}{24})}
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the numerator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }
tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the fraction
tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})
tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}

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Answer:

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Step-by-step explanation:

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