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3 years ago

How do you solve 9 tenths minus 13 fiftenths

Mathematics

1 answer:

Elena L [17]3 years ago

4 0

Answer:

0.04

Step-by-step explanation:

1. 9/10 = 0.9 and 13/15 = 0.86

2. Subtract 0.9 and 0.86

3. You should end up with 0.04

Remember when a number goes like 0.86666666666 to use 0.86 or if it has 0.86666666667 to use 0.87 because you would be simplifying it.

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Vinvika [58]

164×13.7/100=22.47 kg

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torisob [31]

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Step-by-step explanation:

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3 years ago

Simplify the following expression 4 + 3 . 7

Brilliant_brown [7]

Multiply the numbers
4 + 3 . 7
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3 years ago

Question 13 of 20 :

grin007 [14]

Answer:

C. 70.4

Step-by-step explanation:

22% × 320 =

(22 ÷ 100) × 320 =

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3 years ago

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f

zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time $t$

At $t=0, the weight of the bucket is 6+36=42 \mathrm{lb}$

After $t$ seconds, the weight of the bucket is $42-0.15 t \mathrm{lb}$

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after $t$ seconds is $x=1.5 t$

Since the well is 80 ft deep, the time to pull out the bucket is $\frac{80}{2}=40 \mathrm{~s}$

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

$W=\int_{t=0}^{t=36} F d x$

Use $x=1.5 t$ and $F=42-0.15 t$

$W=\int_{0}^{36}(42-0.15 t)(1.5 d t)$

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

$W=3200 \mathrm{ft} \cdot \mathrm{lb}$

4 0

3 years ago