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deff fn [24]
3 years ago
8

Solve the system with elimination. 2x+2y=4 x+3y=6

Mathematics
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

Point Form:

(0,2)

Equation Form:

x=0,y=2

Step-by-step explanation:

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Answer:

2.66 or 2 2/3

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Your answer is 288 dogs

because (216/3 =72, 72 x 4 =288)

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The human eye contains approximately 9.1 x 10^7 rod cells and 4.5 x 10^6 cone cells . Samuel subtracts 9.1-4.5 and says there ar
Paul [167]

Samuel is incorrect because he didn't subtract the power of the numbers.

<h3>How many more rod cells are there?</h3>

The number of rod and cone cells in the human eye is written in scientific notation. Scientific notation is used in expressing large numbers in smaller numbers. To write a number in scientific notation, the number is written as a decimal number, between 1 and 10 and multiplied by a power of 10. For example, 1 x 10² is equivalent to 100

(9.1 x 10^7) - (4.5 x 10^6)

= (9.1 - 4.5) x 10^(7 - 6)

4.6 x 10 = 46

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6 0
1 year ago
What would the graph look like
Alex777 [14]

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Step-by-step explanation:

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2 years ago
Does anyone know the answer?
laila [671]

Answer:

a. θ = 30°

b. μ = √3 / 15 ≈ 0.115

Step-by-step explanation:

Draw a free body diagram for each scenario (see attached figure).  The body has four forces acting on it:

  • Weight pulling down
  • Normal force perpendicular to the incline
  • Applied force parallel up the incline
  • Friction force parallel to the incline

Remember that friction opposes the direction of motion.  So when the body is sliding up, friction points down the incline.  And when the body is sliding down, friction points up the incline.

Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.

For sliding up, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

P₁ − f − mg sin θ = 0

P₁ − Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₁ − mgμ cos θ − mg sin θ = 0

Now for sliding down, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

And sum of the forces parallel to the incline:

∑F = ma

P₂ + f − mg sin θ = 0

P₂ + Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₂ + mgμ cos θ − mg sin θ = 0

We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.

So we have two equations and two unknowns (μ and θ):

P₁ − mgμ cos θ − mg sin θ = 0

P₂ + mgμ cos θ − mg sin θ = 0

Let's start by adding the equations together:

P₁ + P₂ − 2 mg sin θ = 0

P₁ + P₂ = 2 mg sin θ

sin θ = (P₁ + P₂) / (2 mg)

Plugging in the values:

sin θ = (6 + 4) / (2 × 10)

sin θ = 1/2

θ = 30°

Now we can plug this into either equation and find μ.

P₁ − mgμ cos θ − mg sin θ = 0

6 − (10 cos 30°) μ − 10 sin 30° = 0

6 − 5√3 μ − 5 = 0

1 = 5√3 μ

μ = √3 / 15

μ ≈ 0.115

6 0
3 years ago
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