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1 year ago

14

Ms. Rowlen went Black Friday shopping and she bought a flat screen TV that originally costs $259.00. She used her super Savings

Coupon of 45%, how much did Ms. Rowlen pay for the TV?

1 answer:

Paladinen [302]1 year ago

7 0

Answer:

116.55$

Step-by-step explanation:

$259 \times \frac{45}{100} = 116.55$

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the arnold family arrived at the beach at 10:30 a.m. they spent 3¾ hours there.what time did they leave the beach?

JulsSmile [24]

3/4 of an hour is 45 minutes. So they spent 3 hours and 45 minutes at the beach. They arrived at 10:30 so just add 3 hours and 45 minutes to their arrival time and you get 2:15pm. They left at 2:15pm

10:30+3:45=14:15
14:15=2:15

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3 years ago

Evaluate the given expression if x = 25, y = 10, w = 11, and z = 18.<br><br> (see the image)

hammer [34]

Answer:

D

Step-by-step explanation:

So we have the expression:

$(x-y)^2+10wz$

And we want to evaluate it when x=25, y=10, w=11, and z=18.

So, substitute these values for the variables:

$=(25-10)^2+10(11)(18)$

First, subtract within the parentheses:

$=(15)^2+10(11)(18)$

Square 15:

$=225+10(11)(18)$

Now, multiply the terms on the right. 10 times 11 is 110:

$=225+110(18)$

Multiply:

$=225+1980$

Finally, add:

$=2205$

So, our answer is D.

And we're done!

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2 years ago

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Rick has 1/2 of a footlong sub left from yesterday. He ate 1/3 of the leftover sandwich as a snack. What fraction of the entire

Lelechka [254]

The answer is 1/6 of the sandwich

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3 years ago

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

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Answer:

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2 years ago

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