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3 years ago

Lisa is 53 3 4 inches tall. Joey is 1 1 2 inches shorter than Lisa and Jane is 1 1 5 inches shorter than Joey. How tall is Jane?

Mathematics

1 answer:

tester [92]3 years ago

3 0

Answer: Jane's height = $51\dfrac{1}{20}\text{ inches}$

Step-by-step explanation:

Given: Lisa's height = $53\dfrac34\text{ inches}=\dfrac{4\times53+3}{4}\text{ inches}$

$=\dfrac{215}{4}\text{ inches}$

Joey's height = $\dfrac{215}4-1\dfrac12$ inches

$=\dfrac{215}4-\dfrac32\\\\=\dfrac{215-6}{4}\\\\=\dfrac{209}{4}\text{ inches}$

Jane's height = $\dfrac{209}{4}-1\dfrac15$ inches

$=\dfrac{209}{4}-\dfrac{6}{5}\\\\=\dfrac{209\times5-6\times4}{20}\\\\=\dfrac{1045-24}{20}\\\\=\dfrac{1021}{20}\\\\=51\dfrac{1}{20}\text{ inches}$

Hence, Jane's height = $51\dfrac{1}{20}\text{ inches}$

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ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.

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Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

$[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z$

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

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(13)

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