0.096 ÷ 8 and how to get the answer? Thanks!!

Mathematics

2 answers:

Natalka [10]3 years ago

8 0

0.012. (0.096 divided by 8 = 0.012.)

Ksju [112]3 years ago

5 0

When you have 0's in the front like that just ignore them and divide 96/ 8 which equals 12 then add your zeros back and the decimal the correct amount of spaces so its be 0.012 as your answer. hope this helped.

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Item 20 A rectangular school banner has a length of 54 inches and a width of 36 inches. A sign is made that is similar to the sc

siniylev [52]

Answer:

The ratio of the area of the school banner to the area of the sign is <u>1944 cubic inches : 192.61 cubic inches.</u>

Step-by-step explanation:

Given:

A rectangular school banner has a length of 54 inches and a width of 36 inches. A sign is made that is similar to the school banner and has a length of 17 inches.

Now, to find the ratio of the area of the school banner to the area of the sign.

Dimensions of school banner :

Length = 54 inches.

Width = 36 inches.

Dimension of school sign:

Length = 17 inches.

So, to we find the width of sign by using cross multiplication method:

Let the width be $x.$

So, 54 is equivalent to 36.

Thus, 17 is equivalent to $x.$

$\frac{54}{36} =\frac{17}{x}$

By cross multiplying we get:

$54x=612$

Dividing both sides by 54 we get:

$x=11.33\ inches.$

Thus, the width of sign = 11.33 inches.

Now, to get the ratio of the area of the school banner to the area of the sign:

Area of the school banner : Area of the school sign.

= $54\times 36:17\times 11.33$

= $1944:192.61$

Therefore, the ratio of the area of the school banner to the area of the sign is 1944 cubic inches : 192.61 cubic inches.

6 0

3 years ago

Read 2 more answers

If θ=0rad at t=0s, what is the blade's angular position at t=20s

babunello [35]

The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
              ω = 250 (rpm) = 250 * (2π/60) = (25/3)π    rad/s
              ω = 100 (rpm) = 100 * (2π/60) = (10/3)π    rad/s

and from the figure it is clear that the operation is at constant speed but with variable levels
            ⇒   ω = dθ/dt   ⇒   dθ = ω dt

            ∴    θ = ∫₀²⁰ ω dt

while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
       ω = 0                                          from t = 0 to t = 5
       ω = 250 (rpm) = (25/3)π            from t = 5   to t = 15
       ω = 100 (rpm) = (10/3)π            from t = 15 to t = 20

∴ θ = ∫₀⁵ (0) dt   + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt

the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π

∴ θ = 0 + (250/3)π + (50/3)π = 100 π

while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad

Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50