Find the two ordered pairs that are 2.5 units apart

What is the value of the growth factor of the function? A) 1/5 B) 1/3 C) 2 D) 6

ICE Princess25 [194]

Answer:

B

Step-by-step explanation:

Divide 18 by one third and see what ya get

then 6, then 2. BOOM!

7 0

3 years ago

.What is answer pls help me with this very important question. 9 + 10 (78) =

Korolek [52]

Answer:

789

Step-by-step explanation:

9+10(78)

9+10*78

9+780

789

6 0

3 years ago

Read 2 more answers

What are two other names for Xy

NARA [144]

Answer:

In Mathematics, x and y are examples of variables.

You can use any letter as a variable but, make sure to include a let statement when using variables.

A let statement essentially tells what the variable represents.

Ex: let h = houses sold

Hope this helps!

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the lateral area for the pyramid with the equilateral base

likoan [24]

<h3>The lateral area for the pyramid with the equilateral base is 144 square units</h3>

Solution:

The given pyramid has 3 lateral triangular side

The figure is attached below

Base of triangle = 12 unit

Find the perpendicular

By Pythagoras theorem

$hypotenuse^2 = opposite^2 + adjacent^2$

Therefore,

$opposite^2 = 10^2 - 6^2\\\\opposite^2 = 100 - 36\\\\opposite^2 = 64\\\\opposite = 8$

Find the lateral surface area of 1 triangle

$\text{ Area of 1 lateral triangle } = \frac{1}{2} \times opposite \times base$

$\text{ Area of 1 lateral triangle } = \frac{1}{2} \times 8 \times 12\\\\\text{ Area of 1 lateral triangle } = 48$

Thus, lateral surface area of 3 triangle is:

3 x 48 = 144

Thus lateral area for the pyramid with the equilateral base is 144 square units

5 0

2 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