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3 years ago

Find the area of a square with side = 6 cm.

Mathematics

2 answers:

myrzilka [38]3 years ago

4 0

Answer:

isn't it 36cm^2

Step-by-step explanation:

sqaure is even so i think you do 6×6

VikaD [51]3 years ago

4 0

Answer:

36cm

Step-by-step explanation:

Area of a square= Side^2

= 6^2

= 36

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What is the distributive property in -6 (x - 6)-

LenaWriter [7]

Answer:

-6x+36

Step-by-step explanation:

-6 (x - 6)

Distribute

-6x -6 (-6)

-6x+36

8 0

3 years ago

PART 3 OF ME ASKING FOR HELP

e-lub [12.9K]

Answer:

Step-by-step explanation:

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3 years ago

Johanna wrote the system of equations. 4 x - 3 y = 1, 5 x + 4 y = 9. If the second equation is multiplied by 4, what should the

polet [3.4K]

Answer:

The first equation must be multiplied by -5 to eliminate x variable by addition

Step-by-step explanation:

4 x - 3 y = 1 (1)

5 x + 4 y = 9 (2)

If the second equation is multiplied by 4

5x+4y=9. ×4

We have,

20x+16y=36 (3)

The first equation should be multiplied by -5 to eliminate x variable by addition

4x-3y=1 × -5

We have

-20x+15y=-5 (4)

Add equation (3) and (4) to eliminate x variable

20x+16y=36

-20x+15y=-5

31y=31

Divide both sides by 31

y=1

Substitute y=1 into equation (1)

4 x - 3 y = 1

4x-3(1)=1

4x-3=1

4x=1+3

4x=4

Divide both sides by 4

x=1

4 0

3 years ago

Read 2 more answers

HELP Please!!! I need this done quick.

matrenka [14]

What form do you need it in?

6 0

3 years ago

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An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

2 years ago