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Brums [2.3K]
3 years ago
7

Explain how you would distribute 4 ones into 5 groups.m​

Mathematics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

Step-by-step explanation:

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I think of a number and divide it by 3. The result is 2 less than the number I first thought of. Find the number I first thought
Svetach [21]

Answer:

Let the number be x

so, x/4 = x-3

x = 4(x-3)

x= 4x-12

x-4x = -12

-3x = -12

x= 12/3

x = 4

Step-by-step explanation:

Hope this helps! (:

<h3>Please mark me Brainliest!</h3>
6 0
2 years ago
20. Carmen can buy bottles of paint for $2.00 each and boxes of colored pencils for $3.50 each. She can spend no more than
KIM [24]

Answer:

  (5, 8), (7, 1), (15, 3)

Step-by-step explanation:

a) The variables are defined in the problem statement. The total expenditure must satisfy ...

  2.00x +3.50y ≤ 42.00

b) Possible solutions include ...

 (x, y) ∈ {(5, 8), (7, 1), (15, 3)}

__

These were chosen by looking at the attached graph of the equation.

6 0
3 years ago
Please help MeEE!!!!
Dima020 [189]

Answer:

Six inches.

Step-by-step explanation:

1.5 + 2 + 2.5 = 6

6 0
4 years ago
Read 2 more answers
Sorry for the blur but I need help with this one objective: find all seven angles
Natasha_Volkova [10]
I know u need to add diverse then add again then u will get your answer which is 25
3 0
3 years ago
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
4 years ago
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