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Evgesh-ka [11]
2 years ago
6

I need help with systems of equations!! Problem 7-74!

Mathematics
1 answer:
Vika [28.1K]2 years ago
7 0
Problem set A: A=6 and B=2

6=AxB^0 [6=6x1]

24=AxB^2= [24=6x2^2]


Problem set B: A=2 and B=4

32=AxB^2 [32=2x4^2]

128=AxB^3 [128=2x4^3]
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A Plus Plumbing

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60(2)+80 = 120+80 = 200 < 250

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Your answer is 24/36!!!
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Ln(x + 1) − ln(2) = 1
KonstantinChe [14]

Answer:

The answer to your question is: x = 4.44

Step-by-step explanation:

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                                          ln(x + 1) − ln(2) = 1

                                          In (x + 1)/2 = 1

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8 0
3 years ago
A rectangular parcel of land is 210 ft wide. The length of a diagonal between opposite corners is 70 ft more than the length of
Artyom0805 [142]

Width of the rectangular parcel of land = 210 ft.

Let us assume length of the parcel  = x ft.

We are given "the length of a diagonal between opposite corners is 70 ft more than the length of the parcel."

We took x feet for the length of the parcel.

70 ft more than x would be = (x+70).

Diagonal, length and width of the parcel form a right angle triangle, because all angles of a rectangle of 90 degree.

Therefore, we would apply Pythagorean Theorem in that right triangle to find the value of x.

(Width)^2 + (Lengh)^2 = (Diagonal)^2

Plugging values of width, length and diagonal in the above formula.

(210)^2 + (x)^2 = (x+70)^2

44100 + x^2 = x^2 + 4900 + 140x.

Subtracting both sides 4900, we get

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Subtracting x^2 from both sides.

39200 + x^2-x^2  = x^2 +140x-x^2

39200 = 140x

Dividing both sides by 140, we get

39200/140 = 140x/140

x=280 ft.

Therefore, length of the parcel is 280 ft.

4 0
4 years ago
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