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pochemuha
4 years ago
14

The number of Apple cakes is 4/5 of the number of strawberry cakes. There are 24 strawberry cakes more than apple pies. What per

centage of cakes are cranberry? Mr. Smith baked 144 Cakes of cranberry
Mathematics
1 answer:
serious [3.7K]4 years ago
5 0
--------------------------------------------------------------------
Ratio Given
--------------------------------------------------------------------
Apple : Strawberry
4 : 5

--------------------------------------------------------------------
Find difference in parts
--------------------------------------------------------------------
5 - 4 = 1

--------------------------------------------------------------------
Find Number of strawberry cakes
--------------------------------------------------------------------
1 part = 24 cakes
5 parts = 24 x 5 = 120

--------------------------------------------------------------------
Find number of apple cakes
--------------------------------------------------------------------
1 part = 24 cakes
4 parts = 24 x 4 = 96 

--------------------------------------------------------------------
Total number of cakes 
--------------------------------------------------------------------
120 + 96 + 144 = 360

--------------------------------------------------------------------
Percentage of cranberry cakes
--------------------------------------------------------------------
144/360 x 100 = 40%

--------------------------------------------------------------------
Answer: 40% of the cakes are cranberry cakes
--------------------------------------------------------------------
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If y is 5 when x is 2.5 and y varies directly with x, find y when x is 10
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Answer:

If x varies directly as y, then x=ky. y=5 and x=2.5 are substituted into x=ky to find k.

2.5=5k

k=2.5/5

k=5.

K =5 is substituted in x=ky to find y where x is 10.

x=5y

10=5y

y=10/2

y=2

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4 years ago
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Digiron [165]

Answer: The distace between midpoints of AP and QB is \frac{a}{8}.

Step-by-step explanation: Points P and Q are between points A and B and the segment AB measures a, then:

AP + PQ + QB = a

According to the question, AP = 2 PQ = 2QB, so:

PQ = \frac{AP}{2}

QB = \frac{AP}{2}

Substituing:

AP + 2*(\frac{AP}{2}) = a

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Since the distance is between midpoints of AP and QB:

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\frac{AP}{2} = \frac{a}{4}

<u>Midpoint of QB</u>:

\frac{QB}{2} = \frac{a}{4}*\frac{1}{2}

\frac{QB}{2} = \frac{a}{8}

The distance is:

d = \frac{a}{4} - \frac{a}{8}

d = \frac{a}{8}

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Step-by-step explanation:

\large\underline{\sf{Solution-}}

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can be re-arranged as

\rm :\longmapsto\:\dfrac{1}{ \sqrt{3}   -   \sqrt{2}   +  \sqrt{5} }

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }

On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{1}{( \sqrt{3}  -  \sqrt{2} ) +  \sqrt{5} }  \times \dfrac{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }{( \sqrt{3}  -  \sqrt{2} ) -  \sqrt{5} }

We know,

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On rationalizing the denominator, we get

\rm \:  =  \: \dfrac{-  \sqrt{3} +  \sqrt{2}  + \sqrt{5}}{2 \sqrt{6}}  \times \dfrac{ \sqrt{6} }{ \sqrt{6} }

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Hence,

\boxed{\tt{ \rm \dfrac{1}{ \sqrt{3}  +  \sqrt{5}  -  \sqrt{2} } =\dfrac{-  \sqrt{3 \times 3 \times 2} +  \sqrt{2 \times 2 \times 3}  + \sqrt{30}}{12}}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h3><u>More Identities to </u><u>know:</u></h3>

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