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Fofino [41]
3 years ago
9

Which of the following points is a solution to the system of linear inequalities

Mathematics
2 answers:
Nitella [24]3 years ago
4 0

Answer:

(-4,-1) satisfies the given system of inequalities

Step-by-step explanation:

2x+y

-x+y>0

To find out the solution we check with each option

(4,1) , x=4 and y=1 (Plug in x and y values in the given inequalities)

2(4)+1

9 False

(-4,-1) , x=-4 and y=-1 (Plug in x and y values in the given inequalities)

2(-4)-1

-9 True

Now plug it in second inequality

-(-4)-1>0

3>0 True

(-8,-21) , x=-8 and y=-21 (Plug in x and y values in the given inequalities)

2(-8)-21

-37 True

Now plug it in second inequality

-(-8)-21>0

-13>0 False

(8,11) , x=8 and y=-11(Plug in x and y values in the given inequalities)

2(8)+11

-9 false

Sergeu [11.5K]3 years ago
3 0
For it to be a solution, it has to satisfy both inequalities...

subbing in (-4,-1)

2x + y < -5                           -x + y > 0
2(-4) - 1 < - 5                      -(-4) - 1 > 0
-8 - 1 < -5                              4 - 1 > 0
-9 < -5.....true                        3 > 0....true

solution is (-4,-1)

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What is the solution of the equation 9^2x=6? Round your answer to the nearest ten-thousandth.
sammy [17]

Answer:

0.4077

Step-by-step explanation:

It's important to use parenthesis to avoid confusion.

9^(2x) = 6

Use exponent properties:

(9^2)^x = 6

81^x = 6

Take log of both sides

ln(81^x) = ln(6)

Use log properties:

x ln(81) = ln(6)

Solve for x:

x = ln(6) / ln(81)

x ≈ 0.4077

4 0
2 years ago
the perimeter of the floor of a room is 18 metre and its height is 3 cm what is the area of 4 walls of the room​
Lina20 [59]

Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.

Given:

Perimeter of the floor of a room = 18 metre

Height of the room = 3 metre

To find:

The area of 4 walls of the room.

Solution:

We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.

The curved surface area of the cuboid is

C.S.A.=2h(l+b)

Where, h is height, l is length and b is breadth.

Perimeter of the rectangular base is 2(l+b). So,

C.S.A.=\text{Perimeter of the base} \times h

Putting the given values, we get

C.S.A.=18\times 3

C.S.A.=54

Therefore, the area of 4 walls of the room is 54 sq. metres.

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