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3 years ago

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The length of a rectangle is 5 less than twice the width. If the perimeter of the rectangle is 146 cm, find the dimensions of th

e rectangle.
what would the length and width be?

1 answer:

Marta_Voda [28]3 years ago

4 0

Length = 2x - 5

Width = x

Perimeter = 146

P = 2L + 2W

146 = 2(2x - 5) + 2x

146 = 4x - 10 + 2x

146 = 6x - 10

146 + 10 = 6x

156 = 6x

156/6 = x

26 = x

Length = 2(26) - 5

Length = 52 - 5

Length = 47 cm

Width = x

Width = 26 cm

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Lucy has a bag of marbles. She shares one third of them with Carol, keeping one quarter for herself. Amrita joins them, so Lucy

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Step-by-step explanation:

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Lucy shares 1/3 with Carol and keeps 1/4 for herself.

This means that Carol has 1/3 of the total amount of marbles, Lucy has 1/4 of the total amount of marbles.

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We are also told that Carol has 7 more marbles than Lucy

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1/3x = 1/4x + 7

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Answer:

Part (A): The required inequality is T < 35 or t >40.

Part (B): The correct option is C) Only x=7.

Part (C) |x+1|+5=2 has no solution; |4x+12|=0 has one solution; |3x|=9 has two solution.

Step-by-step explanation:

Consider the provided information.

Part (A)

Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer than 40°F is an unsafe food practice.

The union is written as A∪B or “A or B”.

The intersection of two sets is written as A∩B or “A and B”

We need to determine the inequalities represent the union of these improper storage.

That means we will use A∪B or “A or B”.

The improper storage temperatures is when temperature is less than 35°F or greater than 40°F.

Hence, the required inequality is T < 35 or t >40.

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Solve the inequality as shown below:

Subtract 4 from both sides.

$15| x-7|+4-4=10|x-7|+4-4$

Subtract 10|x-7| from both sides

$15\left|x-7\right|-10\left|x-7\right|=10\left|x-7\right|-10\left|x-7\right|$

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Part (C) Match the solution,

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Subtract 2 from both sides.

$\left|x+1\right|=-3$

Absolute value cannot be less than 0.

Hence, |x+1|+5=2 has no solution.

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$4x+12=0$

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Hence, |4x+12|=0 has one solution.

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$\mathrm{If}\:|u|\:=\:a,\:a>0\:\mathrm{then}\:u\:=\:a\:\quad \mathrm{or}\quad \:u\:=\:-a$

$3x=-9\quad \mathrm{or}\quad \:3x=9\\x=-3\quad \mathrm{or}\quad \:x=3$

Hence, |3x|=9 has two solution.

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