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Strike441 [17]
2 years ago
8

Hey y’all can u please give me the correct answer.

Mathematics
1 answer:
ANEK [815]2 years ago
7 0

Step-by-step explanation:

-8 - 6x < 1

Bringing like terms on one side

-6x < 1 + 8

-6x < 9

x < 9/-6

x < 3/-2

I think the correct answer is none

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Draw diagram to rerpresent the mixed number 3 2/3. How can u write this as a single fraction greater than one
muminat
11/3 draw whatever u want
8 0
3 years ago
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Answer of a ²-10a+16-6b-b^2​
german

\quad a^2-10a+16-6b-b^2

$=(a^2-10a)-(b^2+6b) +16$

$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$

$=(a-5)^2-25-(b+3)^2+9+16$

$=(a-5)^2-(b+3)^2$

5 0
3 years ago
A bag contains 5 2/5kg of sand. The bag has a hole in it. Every 10 minutes, 1 2/3kg of sand escapes from the bag. After 20 minut
fenix001 [56]

Quantify of sand in a bag =

= \: 5 \frac{2}{5}  \ \:

=  \frac{(5 \times 5 +2) }{5}

=  \frac{27}{5}  \: kg

Quantity of sand that escapes in every 10 minutes :-

= 1 \frac{2}{3} kg

Quantity of sand that will escape in 20 minutes :-

= 1 \frac{2}{3}  \times 2

=  \frac{(3 \times 1 + 2)}{3}  \times 2

=  \frac{5}{3}  \times 2

=  \frac{10}{3}

Quantity of sand that will remain in the sand bag :-

=  \frac{27}{5}  -  \frac{10}{3}

=  \frac{(27 \times 3)}{15}  -  \frac{(10 \times 5)}{15}

=  \frac{81}{15}  -  \frac{50}{15}

=  \frac{31}{15}

= 2 \frac{1}{5}  \: kg

therefore \:  ,  \: 2 \frac{1}{5}  \: kg \: sand \: is \: left \: in \: the \: bag \: .

3 0
3 years ago
Need help please help me
Ostrovityanka [42]
The solution is the point of intersection between the two equations.

Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.

In the graph the first equation is in blue and the second in red.

The point of intersection = the solution = (-6 , -1)



If you dont have access to a graphing calculator you could draw the graphs by hand;

1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)

y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)

x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3

y = x + 5

x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7

2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph

3) Using a ruler draw a line connection all the dots of one graph and do the same for the other

4) The point of intersection is the solution

8 0
3 years ago
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sattari [20]
63? I just multiplied 9x14/2
8 0
3 years ago
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