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marin [14]
2 years ago
7

Zander read 13 3/4pages in 1/4 of an hour. Sydney read 43 1/3 pages in 2/3 of an hour. At these rates, would Zander or Sydney re

ad more pages in 1 hour?
Mathematics
1 answer:
kakasveta [241]2 years ago
4 0

Answer:

At these rates Sydney read 10 pages per hour more than Zander.

Step-by-step explanation:

Zander read 13\frac{3}{4} pages in \frac{1}{4} of an hour.

Therefore, pages read by Zander in 1 hour = \frac{\text{Total pages read}}{\text{Total time taken}}

= \frac{13\frac{3}{4}}{\frac{1}{4}}

= \frac{55}{4}\times \frac{4}{1}

= 55 pages per hour

Sydney read 43\frac{1}{3} pages in \frac{2}{3} of an hours.

Pages read by Sydney in 1 hour = \frac{43\frac{1}{3}}{\frac{2}{3} }

= \frac{130}{3}\times \frac{3}{2}

= 65 pages per hour

Therefore, number of pages read more by Sydney as compared to Zander,

= 65 - 55

= 10 pages per hour

At these rates Sydney read 10 pages per hour more than Zander.

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If the total number of games was 100, the win amount would need to be 25.

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD=√2 cm.
Zielflug [23.3K]

Given that triangle ABC is right angle triangle. CD is altitude such that AD=BC

ABC is a right angle triangle so apply Pythogorean theorem

AC^{2} + BC^{2} = AB^{2}

AC^{2} + AD^{2} = AB^{2}      (Given that AD = BC)

AC^{2} + AD^{2} = 3^{2}      (Given that AB=3)

AC^{2} + AD^{2} = 9 ...(i)

ADC is a right angle triangle so apply Pythogorean theorem

AD^{2} + CD^{2} = AC^{2}

AD^{2}+(\sqrt{2})^2= AC^{2}

AD^{2}+2= AC^{2}

AD^{2}=AC^{2} -2 ...(ii)


Plug value (ii) into (i)

AC^{2} + AC^{2}-2 = 9

2AC^{2} -2=9

2AC^{2} =11

AC^{2} =\frac{11}{2}

AC=\sqrt{\frac{11}{2} }


Hence final answer is AC=\sqrt{\frac{11}{2} }

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