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Vedmedyk [2.9K]
3 years ago
11

Gỉa sử GDP là 8000$, thuế là 1500$, tiết kiệm tư nhân là 500$, tiết kiệm chính phủ là 200$. Gỉa sử đây là nền kinh tế đóng. Tính

giá trị tiêu dùng
Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0

i dont understand pls tell me in english

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Solve
Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

Ix-4I+3=3

subtracting 3 from both sides (*as per general practice in equations unknown are on one side and constants on the other side or simply taking 3 to other side of the equation)

Ix-4I+3-3=3-3

Ix-4I=0

x-4=0

x-4+4=0=4

x=4

checking:

I4-4I+3=3

I0I+3=3

3=3

8 0
3 years ago
I only need number 10 plz help!!!!!!!
irga5000 [103]

9514 1404 393

Answer:

  • DE ≈ 9.49, EF ≈ 8.06, FD ≈ 14.87
  • obtuse scalene triangle

Step-by-step explanation:

For finding side lengths, it is convenient to work with the differences of the coordinates.

  DE = E(-1, -3) -D(8, -6) = (-9, 3)

  EF = F(-2, 5) -E(-1, -3) = (-1, 8)

  FD = D(8, -6) -F(-2, 5) = (10, -11)

Then the lengths are ...

  DE = √((-9)² +3²) = √90 = 3√10 ≈ 9.49

  EF = √((-1)² +8²) = √65 ≈ 8.06

  FD = √(10² +(-11)²) = √221 ≈ 14.87

The lengths are all different and the largest angle is obviously more than 90°, so the triangle is an obtuse scalene triangle.

4 0
3 years ago
How to find the repeating decimal of 29/13
8_murik_8 [283]
Divide 13 into 29:
2.230769230769....

13 ) 29.0000000
26
—-
3 0 this remainder repeats 6 steps further down
2 6
——
40
39
——
100
91
——
90
78
—
120
11 7
—-
30 which will lead to a recurring decimal because we had remainder 3 at the beginning
6 0
3 years ago
NEED HELP Find values for a, b, c, and d so that the following matrix product equals the 2X2 identity matrix. Explain or show ho
ehidna [41]

We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

<h3>Presenting the equation:</h3>

Basically, we want to solve:

\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]

The matrix product will be:

\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]

Then we must have:

-b + 2 = 1

This means that:

b = 2 - 1 = 1

  • b = 1

We also need to have:

a*b + 1 = 0

we know the value of b, so we just have:

a*1 + b = 0

  • a = -1

Now the two remaining equations are:

-c + 2d = 0

a*c + d = 1

Replacing the value of a we get:

-c + 2d = 0

-c + d = 1

Isolating c in the first equation we get:

c = 2d

Replacing that in the other equation we get:

-(2d) + d = 1

-d = 1

  • d = -1

Then:

c  = 2d = 2*(-1) = -2

  • c = -2

So the values are:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

4 0
3 years ago
Solve.<br> a + 1 = Vb + 1 for b.
Katarina [22]

Answer:

b = a² + 2a

Step-by-step explanation:

a + 1 = √(b + 1)

(a + 1)² = b + 1 (Square both sides)

a² + 2a + 1 = b + 1 (Expand left side)

a² + 2a = b (Subtract 1)

8 0
4 years ago
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