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3 years ago

Please help! Right answer gets brainiest!

Mathematics

2 answers:

solong [7]3 years ago

8 0

Answer:

3^2

Step-by-step explanation:

the answer is 9, so 3^2 is the right answer

bekas [8.4K]3 years ago

4 0

The answer is 3^2.
With multiplication of the powers, we can add them. 3+3=6, so we have 3^6 at the top. We can then put into factor the bottom of the fraction. Since this is division, you can subtract the powers. 6-4=2, so 2 is the final power on the 3.

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Which statement best describes the area of Triangle ABC shown below?

sergeinik [125]

Answer:

It is one-half the area of a rectangle with sides 4 units × 3 units

Step-by-step explanation:

One side of the triangle is on the line y = 2 between points x=2 and x=6. So, that side has length 6-2 = 4.

The opposite vertex has y-value 5, so is 3 units away from the line y = 2.

The area of the triangle can be considered to have a base of 4 and a height of 3. In the formula ...

A = (1/2)bh

we find the area to be ...

A = (1/2)×(4 units)×(3 units) . . . . triangle area

A rectangle's area is the product of its length and width. So, a rectangle that is 4 units by 3 units will have an area of ...

A = (4 units)×(3 units) . . . . rectangle area

Comparing the two area formulas, we see that the triangle area is 1/2 the area of the rectangle with sides 4 units × 3 units.

5 0

4 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

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

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

Suppose S is between R and T. Use the Segment Addition Postulate to solve for z.. RS=2z+6, ST=4z−3, RT=5z+12

Lynna [10]

Answer:

Step-by-step explanation:

If S is between R and T, then RS+ST = RT

Given the following parameters

RS=2z+6,

ST=4z−3,

RT=5z+12

On substituting

2z+6+4z-3 = 5z+12

Collect like terms

2z+4z-5z = 12+3-6

6z-5z = 15-6

z = 9

Hence the value of z is 9

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4 years ago

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6 0

3 years ago

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BlackZzzverrR [31]

Answer:

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Step-by-step explanation:

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4 years ago