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2 years ago

Which of the following is equivalent to 60 1/2

Mathematics

1 answer:

kotykmax [81]2 years ago

3 0

Answer:

30 or 120 /4

Step-by-step explanation:

a^p/q here p denotes power raised and q denotes inverse power or root. So, 60^1/2 denotes square root of 60 raised to power 1.

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Does this table represent a function? Why or why not?

arsen [322]

Answer:

Step-by-step explanation:

Because their are more than one ranges

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3 years ago

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Suppose FLX =ZAQ Which other congruency statements are correct

Stella [2.4K]

LFX=AZQ
XLF=QAZ
FXL=ZQA
XFL=QZA

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3 years ago

Graph this line using the slope and y-intercept: y = 1/8x + 4<br> (1/8 is a fraction)

Tems11 [23]

Answer:

Two points, (0,4) & (8,5)

Step-by-step explanation:

In this equation, we can easily find the first point by looking at the +4 in this equation. This should automatically signal to you that the point is (0,4)

Now, we have to look at the 1/8x in the equation.

This 1/8 means that the slope is 1/8!

Now we use the slope and to rise/run to find our next point

We go up 1 from 4 (this is the rise) to get to 5 on the y-axis

Next, we go over 8 horizontally to get to 8 for your x-axis point.

So therefore, our next point is, (8,5)

You can use this rise/run system infinitely and is very useful

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2 years ago

What point names the ordered pair (3, -5)?

gregori [183]

Answer:

Step-by-step explanation:

We start at the origin (we the axis cross)

We go three units to the right, then 5 units down.

We end up at the point D

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3 years ago

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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