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sattari [20]
2 years ago
5

Determine the percentage of the square that belongs to each of the four regions. (image included)

Mathematics
1 answer:
tatyana61 [14]2 years ago
3 0

Step-by-step explanation:

maybe you can try to measure each of the ways maybe and you can get the equation

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To run 12 miles, how many times must you run around a park that is 2 miles long and 1 mile wide?
Veronika [31]
Pretty sure it’s 6 miles
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What is the product? (9t − 4)(−9t − 4) −81t2 − 16 −81t2 16 −81t2 − 72t 16 −81t2 72t 16
Alexxandr [17]
<span>We have to find the product : ( 9 t - 4 ) ( - 9 t - 4 ) = - ( 9 t - 4 ) ( 9 t + 4 ) After that we can use the difference of squares. The formula is: ( a + b ) ( a - b ) = a^2 - b^2. In this case: - ( 9 t - 4 ) ( 9 t + 4 ) = - ( 81 t^2 - 16 ) = - 81 t^2 + 16 . Answer: B ) - 81 t^2 + 16.</span>
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3 years ago
1/4 + 5/8 + 1/8 + 1/4 + 1/4 + 1/4 + 3/4 + 3/4
zysi [14]
1/4=2/8  and 3/4=6/8... common denominator 
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7 0
3 years ago
Read 2 more answers
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
2 years ago
Suppose the half-life of a certain radioactive substance is 12 days. Find the time when there will be 20% of the initial substan
vazorg [7]

Answer: 2.4% I think

12 x 20 / 100 = 2.4%

7 0
3 years ago
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