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3 years ago

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an astronaut who weighs 126lb on earth weighs 21lb on the moon. How much would a person who weighs 25lb on the moon weigh on ear

th?

1 answer:

irakobra [83]3 years ago

7 0

If 21 Ib on moon=126 IB on earth
25 Ib on moon=?
25/21*126=150
Therefore the astronaut will weigh 150Ib on earth

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Walden’s family is shopping for a reclining chair. The chair the family decided on has a retail price of $800 plus 5% sales tax

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Answer:

B 10% off sale, with a total price to pay of 756

Step-by-step explanation:

Retail price (Rp) = $800

Taxes (T) = 5%

Offer A $75 instant rebate

Total price A (TA) = Rp - 75 + (Rp - 75) 5%

TA = 800 - 75 + (800-75) 5%

TA = 725 + 725 5%

725 5% = 725*5/100 = 36.25

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Total price B (TB) = RP - RP 10% + (RP - RP 10%) 5%

TB = 800 - 800 10% + (800 - 800 10%) 5%

800 10% = 800*10/100 = 80

TB = 800 - 80 + (800 - 80) 5%

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720 5% = 720*5/100 = 36

TB = 720 + 36 = $756

Offer C 5% off sale plus store pays sales tax

store pays sales tax means Walden's family will not need pay for taxes

Total price C (TC) = RP - RP 5%

TC = 800 - 800 5%

800 5% = 800*5/100 = 40

TC = 800 - 40 = $760

Offer D a “no tax” sale—the store pays the tax

Means that Walden's family will need pay just the retail price

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$\longrightarrow{\mathfrak{\frac{-9}{6}÷\frac{3}{-2}}}$

$\longrightarrow{\sf{\frac{-\cancel{9}}{\cancel{6}}\times\frac{-\cancel{2}}{\cancel{3}}}}$

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$\longrightarrow{\boxed{\bf{1}}}$

Identity applied -

$\star{\:\:\:\:\:\:\boxed{\bf{\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}}}}$

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(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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