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3 years ago

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Mathematics

1 answer:

cupoosta [38]3 years ago

7 0

See above information

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Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

GREYUIT [131]

Answer:

Decay
Decrease rate is 8%

Step-by-step explanation:

Given function:

y = 490(0.92)^x

The base of the exponent is 0.92. This represents a decay as 0.92 is less than 1. It would be a growth function is the base was greater than 1.

The rate of decrease is

0.92 times

Percent decrease is

(1 - 0.92)*100% = 8%

7 0

3 years ago

What is the volume of a sphere with a radius of 6 inches Not pi

Nimfa-mama [501]

Answer:

≈

904.78

in³

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6 0

3 years ago

Seventeen candidates have filed for the upcoming county council election. 7 are women and 10 are men a) Is how many ways can 10

tankabanditka [31]

Answer: (a) 19448 ways

(b) 5292 ways

(c) 0.2721

Step-by-step explanation:

(a) 10 county council members be randomly elected out of the 17 candidates in the following ways:

= $^{n}C_{r}$

= $^{17}C_{10}$

= $\frac{17!}{10!7!}$

= 19448 ways

(b) 10 county council members be randomly elected from 17 candidates if 5 must be women and 5 must be men in the following ways:

we know that there are 7 women and 10 men in total, so

= $^{7}C_{5}$ × $^{10}C_{5}$

= $\frac{7!}{5!2!}$ × $\frac{10!}{5!5!}$

= 21 × 252

= 5292 ways

= $\frac{ ^{7}C_{5} * ^{10}C_{5}}{^{17}C_{10}}$

= $\frac{5292}{19448}$

= 0.2721

8 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago

Please help me ☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️

Olegator [25]

Two complementary angles make a total of 90, so 3x+6x=90, x=10

5 0

3 years ago

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