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3 years ago

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Two points on the graph of a function are used to determine the average rate of change over the interval between them. Calculate

the average rate of change for the two given points. A. -1/3 B. -2/4 C. 1/3 D. 6/2

1 answer:

dlinn [17]3 years ago

3 0

Answer:

Step-by-step explanation:

lololo

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What is LCM of two numbers if one number is a multiple of the other?Give on Example.

miv72 [106K]

Answer:

the greater number will be the lcm if the other number is a multiple

Step-by-step explanation:

for example: the lcm of 2 and 4

the lcm is 4 as 2 and 4 are both multiples of 4

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3 years ago

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Sweatshirts at the discount store are on sale for 30% off the original price. If a sweatshirt

kobusy [5.1K]

A.) the discount is $9
B.) the new price is $21

Work:
30% is turned to .30
so, $30 x .30 is $9.
that’s how much is taken off.

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3 years ago

The measure of angle a 73 degrees classify the angle as acute right obtuse or straight

aliya0001 [1]

The answer is Acute Angle

4 0

3 years ago

The given two-parameter family is a solution of the indicated differential equation on the interval

mr_godi [17]

Answer:

I do 1 option for you as an example, you need to check the leftover by yourself.

Step-by-step explanation:

for d) y(0) = 0 and y'(pi) =0

$y(0) = C_1e^0cos(0)+ C_2 e^0 sin(0) = 0 \longrightarrow C_1 = 0$

$y(x) ' = C_1e^x cos(x) - C_1e^x sin (x) + C_2e^x sin(x) + C_2e^x cos(x)$

$y(\pi)'=C_1e^\pi cos(\pi)- C_1e^\pi sin(\pi)+ C_2e^\pi sin(\pi) + C_2e^\pi cos (\pi)$

Replace $C_1 = 0$ we have

$y'(\pi) = -C_2e^\pi = 0$

if and only if $C_2 =0$

Hence the given solution does not work.

then, d is NOT the correct answer.

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3 years ago

The matrix C=[1,-2_-3,7] was used to encode a phrase to [7,-28,-25,-35,-2_-21,107,90,123,17]. Find C^-1 and use it to decode the

posledela

$C= \left[\begin{array}{cc}1&-2\\-3&7\end{array}\right] \\ C^{-1}= \left[\begin{array}{cc}7&2\\3&1\end{array}\right] \\ \left[\begin{array}{cc}7&2\\3&1\end{array}\right] \times \left[\begin{array}{ccccc}7&-28&-25&-35&-2\\-21&107&90&123&17\end{array}\right] \\ =\left[\begin{array}{ccccc}49-42&-196+214&-175+180&-245+246&-14+34\\21-21&-84+107&-75+90&-105+123&-6+17\end{array}\right] \\ =\left[\begin{array}{ccccc}7&18&5&1&20\\0&23&15&18&11\end{array}\right]$
is the required matrix.

6 0

3 years ago