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Nataly [62]
2 years ago
5

Fill in the blanks for this geometry problem. Translation, reflection, etc type of problem. 70 POINTS!

Mathematics
1 answer:
Shkiper50 [21]2 years ago
7 0

The answer is translation of 2 units left and a reflection over the x-axis.

The triangle is translated 2 units left because the triangle moved two units to the left and it is reflected over the x-axis because the bottom triangle is like a reflection of the original triangle.

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(-3)(8+-10)!!!!!!!!!!!!!!!!!!!!!
evablogger [386]
(-3)(8+-10)
(-3) • ( -2)
which equals 6
8 0
3 years ago
How do you put 2.303,2.3,2.33 and 2.03 from least to greatest
ser-zykov [4K]
You add extra zeroes until all numbers have the same quantity. in this case, 2.303 has three numbers after the decimal, so the other three number need it too. 2.300, 2.330, and 2.030. now that you have all of them looking the same now, you can tell which is least and work from there: 2.030 is the least, then 2.300, 2.303 then 2.330 is the largest.
8 0
2 years ago
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What is the leading coefficient of the following expression: 5x^4-7x^3+2x+1​
never [62]

Answer:

5

Step-by-step explanation:

The leading coefficient belongs to the term with the largest exponent

Given

5x^{4} - 7x³ + 2x + 1 ← in standard form

The leading term is 5x^{4} ← with leading coefficient 5

5 0
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
2 years ago
1.5 : 2.5<br> how to simplify it
nataly862011 [7]
If you use the common denominator of .5 for both sides of the ratio, you will arrive at 3:5 as a simplified form.
7 0
2 years ago
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