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2 years ago

To be called the Order of the Maandor of minors nonzero in a Matrix

Mathematics

1 answer:

MissTica2 years ago

4 0

Answer:

To be called the order of the matador of Minors nonzero in a matrix

C D Range of a matrix

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Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is

solong [7]

Step-by-step explanation:

The explanation and the graphical representation is attached

6 0

3 years ago

Convert this decimal into its fractional<br> form, simplified completely.<br> 0.650

podryga [215]

Answer:

13/20

Step-by-step explanation:

0.650 /1

0.650 × 100

1 × 100 =

65 /100

65 ÷ 5

100 ÷ 5 =

13 /20

4 0

2 years ago

ANSWER ASAPPP PLSSS. ILL MARK BRAINLIEST TOO

nevsk [136]

Answer:

Step-by-step explanation:

11.04 = 10(1.02)^n

1.104 = 1.02^n

ln 1.104 = ln 1.02^n

ln 1.104 = n ln 1.02

n = ln 1.104/ ln 1.02

n = 4.99630409516

4.99 can be rounded to 5.

So a reasonable domain would be 0 ≤ x < 5

PART B)

f(0) = 10(1.02)^0

f(0) = 10(1)

f(0) = 10

The y-intercept represents the height of the plant when they began the experiment.

f(1) = 10(1.02)^1

f(1) = 10(1.02)

f(1) = 10.2

(1, 10.2)

f(5) = 10(1.02)^5

f(5) = 10(1.1040808)

f(5) = 11.040808

f(1)=10(1.02)^1

f(1)=10.2

Average rate= (fn2-fn1)/(n2-n1)

=11.04-10.2/(5-1)

=0.22

the average rate of change of the function f(n) from n = 1 to n = 5 is 0.22.

3 0

2 years ago

Read 2 more answers

Katrina is asked to simplify the expression .

Lorico [155]

The property of distribution

8 0

3 years ago

Read 2 more answers

In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x

Alexus [3.1K]

Answer:

Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

Step-by-step explanation:

Given Differential Equation,

$y"-2y'+y=xe^x$ ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

$\implies\:y(x)=y_c(x)+y_p(x)$

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

$y_c(x)=C_1e^x+c_2xe^x$

Now for particular solution,

let $y_p(x)=Ax^3e^x$

$y'=Ax^3e^x+3Ax^2e^x$

$y"=Ax^3e^x+6Ax^2e^x+6Axe^x$

Now putting these values in (1), we get

$Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x$

$Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x$

$6Axe^x=xe^x$

$6A=1$

$A=\frac{1}{6}$

$\implies\:y_p(x)=\frac{x^3e^x}{6}$

Therefore, Solution is $y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}$

8 0

3 years ago