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jek_recluse [69]
3 years ago
15

What is the average rate of change for this exponential function for the interval from x=1 x=3

Mathematics
2 answers:
mojhsa [17]3 years ago
6 0
Is there any y values? Or answers that go with this question?
inessss [21]3 years ago
5 0

Answer:


Step-by-step explanation:


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Ross took a survey of his classmates' preferred food as well as recording their genders. The results are in the table below:
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The answer to your question is 30
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3 years ago
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Find the values of x and y that make the quadrilateral a parallelogram
lana66690 [7]

For given parallelogram, the value of x = 114° and y = 66°

<h3>Explanation: </h3>

<em>(Refer figure)</em>

Given quadrilateral ABCD is a parallelogram.

m∠A = 66°

m∠B = x°

m∠C = y°

m∠D = 114°

Since by properties of parallelogram, the both pairs of opposite sides of a parallelogram are parallel.

Opposite angles of a parallelogram are congruent ...........................(1)

Consecutive angles of a parallelogram are supplementary ................(2)

Using (1) for given parallelogram ABCD; we get,

m∠B = m∠D

m∠A = m∠C

Therefore x = 114° and y = 66°.

Using (2) to check whether the two consecutive angles are supplementary.

( m∠A + m∠D ) =  ( m∠B + m∠C ) = ( m∠C + m∠D ) = 66° + 114° = 180°

6 0
3 years ago
The model is made out of a stack of plywood sheets.Each sheet is 0.6 thick.How many sheets of plywood tall is the model?
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at an aquarium six out of 18 deliveries are plants out of 15 delivery's in one week how manybare plants
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Do six multiply fifteen which equals to 90
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6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1 ...?
jarptica [38.1K]
I will use the letter x instead of theta.

Then the problem is, given sec(x) + tan(x) = P, show that

sin(x) = [P^2 - 1] / [P^2 + 1]

I am going to take a non regular path.

First, develop a little the left side of the first equation:

sec(x) + tan(x) = 1 / cos(x) +  sin(x) / cos(x) = [1 + sin(x)] / cos(x)

and that is equal to P.

Second, develop the rigth side of the second equation:

[p^2 - 1] / [p^2 + 1] =

= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =

=  { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =

= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }

= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =

= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =

= sin(x)

Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.


4 0
4 years ago
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