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melomori [17]
2 years ago
7

Which equation is equivalent to 16 Superscript 2 p Baseline = 32 Superscript p 3?.

Mathematics
1 answer:
Mrac [35]2 years ago
7 0

The equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

2^{8p}=2^{5p+15}

<h3>What is equivalent equation?</h3>

Equivalent equation are the expression whose result is equal to the original expression, but the way of representation is different.

Given information-

The given equation in the problem is,

16^{2p}=32^{p+3}

Write both the equation in the form of same base number as,

(2^4)^{2p}=(2^5)^{p+3}

The power of the power of a number can be written as product of both the numbers. Thus,

(2)^{4\times2p}=(2)^{5\times(p+3)}\\2^{8P}=2^{5P+15}

This is the required equation.

Now if the base is the same at both side of the expression, then the powers can be compared. Thus,

8p=5p+15

Solve it further to find the value of p as,

8p-5p=15\\3p=15\\p=5

Thus the equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

2^{8p}=2^{5p+15}

Learn more about the equivalent expression here;

brainly.com/question/2972832

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2 years ago
Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer.
Sonja [21]
<h3>Answer:  Rhombus</h3>

======================================================

Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231\\\\

Segment AB is exactly \sqrt{17} units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

-------------------------

Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2)  = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}\\\\

Segment AB has a slope of 1/4.

Do the same for BC

B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2)  = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4\\\\

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a <u>rhombus</u>.

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12 centimeters blank 98 millimeters
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What's the question?
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