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Nutka1998 [239]
3 years ago
14

—2+7+(-4). Evaluate

Mathematics
2 answers:
MakcuM [25]3 years ago
5 0

Answer:

1

Step-by-step explanation:

-2 + (-4) = -6 + 7 = 1

evaluate = solve

ruslelena [56]3 years ago
4 0

Answer:

The answer is 1

Step-by-step explanation:

-2+7=5

5+(-4)= 1

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ozzi
You do 

-26
-4
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5n=-30 so...

-30/5=-6

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n=-6
3 0
3 years ago
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What is 6 divided by 39?
melomori [17]
6 with a remander of 3 
4 0
3 years ago
Solve for x & y solve x = 3 + yy = -2x + 9
sergey [27]

x = 3 + y Eqn(1)

y = -2x + 9​ Eqn(2)

Let us solve the system of equations with the substitution method

x - 3 = y (Subtracting 3 from both sides of the Eqn(1))

Replacing y = x - 3 in Eqn (2), we have:

x - 3 = -2x + 9

x = -2x + 9 + 3 (Adding 3 to both sides of the equation)

x + 2x = 9 + 3 (Adding 2x to both sides of the equation)

3x = 12 ( Adding like terms)

x = 12/3 (Dividing by 3 on both sides of the equation)

x = 4

Replacing x=4 in Eqn(1), we have:

4 = 3 + y

4 - 3 = y (Subtracting 3 from both sides of the equation)

y=1

The answers are:

x= 4 and y=1

6 0
1 year ago
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
bekas [8.4K]

9514 1404 393

Answer:

  • (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
  • NOT linearly independent

Step-by-step explanation:

We want ...

  c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

  c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

  x(c1 +2c3) = 0

  x²(c2 -4c3) = 0

These have an infinite set of solutions:

  c1 = -2c3

  c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

  (c1, c2, c3) = (-2t, 4t, t)

__

f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0
2 years ago
What is the volume of this right rectangular prism?
vfiekz [6]
The volume of this right rectangular prism is 263.52 and it is to the nearest hundredths as a decimal, hope it helps!
8 0
3 years ago
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