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prohojiy [21]
3 years ago
8

4. Gabby makes banana bread at her bakery every Monday and Wednesday. Her

Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

Top row: 3 cups, 30 tablespoons, 12 bananas

Second row: 7 cups, 70 tbs, 28 bananas

Step-by-step explanation:

The ratio is 5 (sugar) : 50 (butter)...

You can make the unit rate: 1 cup of sugar: 10 tbs of butter

Since the other ratio is 80 tbs : 32 bananas...

1 tbs : .4 bananas

So...

3 cups : 30 tbs

30 tbs : 12 bananas

28 bananas : 70 tbs

70 tbs :  7 cups

Hope this helps!!!

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To find this, simply subtract 57 from 90. 90 is the measurement for a completed compliment equation, and 57 is the part of the compliment we know.

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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
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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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