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Anit [1.1K]
3 years ago
5

Omar’s classroom has 2 closets. Each closet has 3 shelves. There are 5 backpacks on each shelf. How many backs are there in all

Mathematics
1 answer:
almond37 [142]3 years ago
5 0

I believe there are 35 backpacks in all. 2 closets have 6 shelves all together. Then 6*5=35.

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Anyone understand how to do both of these?
Evgen [1.6K]

for the first one set 3y+y=180 and y+x=180


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6 0
3 years ago
Paula is located at a position of −12 feet relative to the ground. Which statement shows the distance she is from ground level?
Greeley [361]

The absolute value of a number is the distance from point to the origin. Since Paula is located at a position of −12 feet relative to the ground, then the distance from point -12 on the number line to the origin is 12 (distance cannot be negative).

Thus, |-12|=12 and Paula is at a distance of 12 feet.

Answer: correct choice is C

3 0
3 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
Nadine is constructing a line perpendicular to KE←→ . She has constructed the arcs as shown. What is the next step in her constr
Margaret [11]
The correct answer is "Draw a straight line from D to M". In order to draw a perpendicular line to line KDE you need to create two acrs that intersect above or below KDE. This is illustrated by the point M. This is the point that will create a perpendicular line and intersect KDE at a 90 degree angle. 
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valkas [14]
Because the time right at launch equals 0 (0 seconds have passed since the launch) solve the equation for 0.
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-5(1)(-9)
-5(-9)
45
The object was 45 metres above the ground at the time of the launch.
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