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Vedmedyk [2.9K]
3 years ago
13

[02.05] Marcie solved the following inequality, and her work is shown below: −2(x − 5) − 12 ≤ 4 + 6(x + 3) −2x + 10 − 12 ≤ 4 + 6

x + 18 −2x − 2 ≤ 6x + 22 −8x ≤ 24 x ≤ −3 What mistake did Marcie make in solving the inequality?
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
4 0
Hi there! Marcie made a mistake in the last step of the solving process.

Let's solve this problem step by step!
−2(x − 5) − 12 ≤ 4 + 6(x + 3)

First work out the parenthesis, which can for instance be done using rainbow method.

−2x + 10 − 12 ≤ 4 + 6x + 18
The parenthesis are worked out correctly. Remember that multiplying a negative by a negative gives you a positive.

Now we can collect the terms.
−2x − 2 ≤ 6x + 22

The terms are collected correctly.

Next up is subtracting 6x from both sides and adding 2 to both sides of the inequality.
−8x ≤ 24

This step is also correct.
Finally we must divide both sides of the inequality by -8 to find our answer. Be aware of the fact that the sign flips when dividing by a negative!

x ≤ −3
Mistake! Marcie forgot to flip the sign. Instead the answer should be x \geqslant - 3

~ Hope this helps you!
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What is the rate of the boat in still water and what is the rate of the current?
Alekssandra [29.7K]

Answer: 2mp

Step-by-step explanation: Let "b" represent the speed of the boat in still water

Let "c" represent the speed of the current

Let "t" represent time

Downstream(same direction) =(b+c)*t

Upstream(against current)=(b-c)*t

(b+c)*3=24

(b-c)*4=16

3b+3c=24....all sides can be divided by 3 =b+c=8

4b-4c=16.....all sides can be divided by 4 =b-c=4

Use Elimination method

b+c=8

b-c= 4 Subtracting

=====

2c=4

find c by dividing both sides by 2. c=2

if c=2, substitute to get b

b+2=8

 -2=-2

======

b=6

Speed of boat in still water =6mph

rate of current=2mp

4 0
2 years ago
Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro
Neporo4naja [7]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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Answer:35in.

Step-by-step explanation:

I reccomend splitting it into sections when given problems like this.

The smaller two rectangles would be 5in * 2in which is 10in then just add them together.

For the bigger one, you'd do 5-2 in order to get the missing inches which will be 3+2 once added 5in*3in is 15in.

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