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kramer
4 years ago
12

Please help me!i have problems with this question

Mathematics
1 answer:
Leno4ka [110]4 years ago
4 0

Answer:

the mass of the empty box is 4/3 kg.

Step-by-step explanation:

The total mass (box plus 4 books and 2 files) is 2 kg

When half of the load is removed (2 books and 2 files) the mass is 1 3/20.

We have the following equation:

2 kg - (x/2)= 1 3/20 kg

Where x is the total mass of the load (4 books and 2 files)

This steps where followed to solve the equation:

2 -x/2 = (20-3)/20

40-10x=17

23=10x

x=2/3 kg

That is the total mass of the load

Hence

2kg -2/3 kg= 4/3 kg, the mass of the box

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3 (5 + 6)
ELEN [110]

Hi there!

Distributive property are two big words that basically just mean you need to distribute the 3 to the whole parentheses. This means you need to multiply 5 by 3 and also 6 by 3.

3 × 5 = 15

3 × 6 = 18


Now that you've distributed the 3 to the entire parentheses, you just need to add together both answers you got :

15 + 18 = 33


Your answer is: 33


There you go! I really hope this helped, if there's anything just let me know! :)

4 0
3 years ago
8x + 11 + 3x + 11 + 39 = 180
kow [346]

Answer:


Step-by-step explanation:

collect terms

11x+61=180

move numbers to one side and x to other

11x=180-61

11x=119

x=119/11=10.8

7 0
3 years ago
Read 2 more answers
Jk kl and lj are all tangent to o ja = 14 al = 15 and ck=13 find the perimerter of jkl
alexgriva [62]
 JK,KL and LJ are all to O.JA =14, AL=15, and CK=13. Find the perimeter of triangle JKL.? The answer is D. 84

6 0
3 years ago
Read 2 more answers
Which description compares the vertical asymptote(s) of Function A and Function B correctly?
r-ruslan [8.4K]
Function A:  f(x)=\frac{1}{x}+4.  Vertical asymptotes are in the form x=, and they are a vertical line that the function approaches but never hits.  They can be easily found by looking for values of <em>x</em> that can not be graphed.  In this case, <em>x</em> cannot equal 0, as we cannot divide by 0.  Therefore <em>x</em>=0 is a vertical asymptote for this function.  The horizontal asymptote is in the form <em>y</em>=, and is a horizontal line that the function approaches but never hits.  It can be found by finding the limit of the function.  In this case, as <em>x</em> increases, 1/<em>x</em> gets closer and closer to 0.  As that part of the function gets closer to 0, the overall function gets closer to 0+4 or 4.  Thus y=4 would be the horizontal asymptote for function A.
Function B:  From the graph we can see that the function approaches the line x=2 but never hits.  This is the vertical asymptote.  We can also see from the graph that the function approaches the line x=1 but never hits.  This is the horizontal asymptote.
3 0
3 years ago
Read 2 more answers
In a journey of 15 miles, two third distance was travelled with 40 mph and remaining with 60 mph. How much time does the journey
Ad libitum [116K]

The first step in solving this question is to split the journey into 2 parts. In the first part of the journey, \frac{2}{3} \times 15 = 10 miles are covered at a speed of 40 mph. In the second part the journey 5 miles are covered at a speed of 60 mph.

The equation to compute the time of a journey  given the speed and distance is t=\frac{d}{s} where t is the time, s is the speed and d is the distance.

The time for the first part of the journey is calculated as shown below,

t=\frac{d}{t} \\t=\frac{10m}{40mph} =\frac{10}{40} h.

The time for the second part of the journey is calculated as follows,

t=\frac{d}{t} \\t=\frac{5m}{60mph} =\frac{5}{60} h.

The total time is the sum of the times taken to cover each part of the journey and is calculated as shown below,

t=\frac{10}{40} h+\frac{5}{60} =\frac{30+10}{120} h=\frac{40}{120}h =\frac{1}{3} h

The time to cover the journey is a third of an hour or 20 minutes.

5 0
3 years ago
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