All categories

2 years ago

Please please help me again, I really need help and thank you.

Mathematics

1 answer:

ipn [44]2 years ago

5 0

Answer:

Perimeter of Compound Shapes: Composite Rectangles. Perimeter is the total distance around the outside of a shape. horizontal. sides must be the same length in total as the one longer red side.

Step-by-step explanation:Also not sure...Hope this kinda helps...

You might be interested in

Isabelle is mixing red paint with blue paint to make purple paint. She adds 3/10 of a fluid ounce of red to 11/15 of a fluid oun

MAXImum [283]

40\13 that what she needs

6 0

2 years ago

Read 2 more answers

Joanne has a scale drawing of her backyard that includes a garden bed that measures 26 inches long and 16 inches wide.

SIZIF [17.4K]

1, 12 2.16 it should be right because i got the same question before

4 0

2 years ago

Can someone help me find the rule?

frutty [35]

I think the rule might be x-9

4 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

2 years ago

Which two triangles are congruent? Complete the congruence statement.

Paladinen [302]

answer:

△DEF ≅ △VTU

step-by-step explanation:

when you compare the figures, you will see how △DEF and △VTU look very similar, while the other one looks different in size
another thing is that they have two congruent sides, and one congruent angle which fulfills the SAS triangle congruence theorem

4 0

2 years ago