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maria [59]
3 years ago
8

There are 80 sixth graders at Wilson Middle School. Only 60% of the sixth graders will attend the morning assembly. How many six

th graders will be at the morning assembly?
help please
Mathematics
2 answers:
iragen [17]3 years ago
6 0
80 x 0.60=48
So what you do is you multiply 80  times 60% which gives you 48 which is that's how many students will attend the assembly in the morning.
Montano1993 [528]3 years ago
3 0
You would do 80 * 6/10, which equals 48. Therefore there would be 48 sixth graders at the morning assembly. Hope this helps. Please rate, leave a thanks, and mark a brainliest answer(Not necessarily mine). Thanks, it really helps! :D
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A student throws a rock into the sky from the top of the Webster build- ing with an angle α = π 4 from the horizontal line and t
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  • r(0) = <0, 100> . . . . . . . .meters
  • r'(0) = <7.071, 7.071> . . . . meters per second

Step-by-step explanation:

<u>Initial Position</u>

The problem statement tells us we're measuring position from the ground at the base of the building where the projectile was launched. The initial horizontal position is presumed to be zero. The initial vertical position is said to be 100 meters from the ground, so (in meters) ...

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<u>Initial Velocity</u>

The velocity vector resolves into components in the horizontal direction and the vertical direction. For angle α from the horizontal, the horizontal component of velocity is v₁·cos(α), and the vertical component is v₁·sin(α). For v₁ = 10 m/s and α = π/4, the initial velocity vector (in m/s) is ...

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7 0
3 years ago
The volume of a rectangular box with a square base remains constant at 500 cm3 as the area of the base increases at a rate of 6
Andre45 [30]

Answer:

the rate at which the height of the box is decreasing is -0.0593 cm/s

Step-by-step explanation:

Given the data in the question;

Constant Volume of a rectangular box with a square base = 500 cm³

area of the base increases at a rate of 6 cm²/sec

so change in the area of the base with respect to time dA/dt = 6 cm²/sec

each side of the base is 15 cm long

so Area of the base = 15 cm × 15 cm = 225 cm²

the rate at which the height of the box is decreasing = ?

Now,

V = Ah

dv/dt = 0 ⇒ Adh/dt + hdA/DT = 0

⇒ dh/dt = -hdA/dt / A

we substitute

dh/dt = [ -( 500 / 225 ) × 6 ] / 225

dh/dt = [ -(2.22222 × 6)  ] / 225

dh/dt = [ -13.3333 ] / 225

dh/dt = -0.0593 cm/s

Therefore, the rate at which the height of the box is decreasing is -0.0593 cm/s

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