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3 years ago

Triangle MAT is similar to Triangle RUG. Complete the statement: Line segment

Mathematics

2 answers:

jok3333 [9.3K]3 years ago

8 0

The answer would be RG because corresponding means they’re in the same position

Oduvanchick [21]3 years ago

3 0

It would be corresponding to RG.

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Find the distance between the points (-4,-1) and (0,-2).

goldfiish [28.3K]

Answer:

-4,-1

Step-by-step explanation:

So how I did it is, I made a mintale picture of it in my head. Then, I "counted" how much space I had to go for the next point. And that's how I got the answer of -4,-1.

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3 years ago

If a new car has 2 transmission types, 3 vehicle styles, 3 option packages, 7 exterior color choices, and 3 interior color

VLD [36.1K]

The number of different new cars that are possible is found as follows:
$2\times 3\times3 \times7\times 3=?$

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3 years ago

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I need help on my homework Number 8. Part A and Part B

Damm [24]

Number 8 is 300 because 10 times 3 equal 300

6 0

3 years ago

Line

mel-nik [20]

Answer:

x=26

Step-by-step explanation:

we know that

m∠KOL+m∠LOM+256°=360° -----> by complete circle

substitute the given values and solve for x

$(x+3x)\°+256\°=360\°\\4x=360-256\\4x=104\\x=26$

5 0

3 years ago

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago