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3 years ago

ANSWER NOW HURRY

Mathematics

2 answers:

Annette [7]3 years ago

5 0

The answer is C because it’s twice as long

stiv31 [10]3 years ago

4 0

Answer:

B It is the same length.

Step-by-step explanation:

Due to transformations, If you both rotate and reflect a segment it will still remain the same...

R' and T' is created if you do either on a coordinate plane.

Working on it on paper:

Lets create a problem...

"Triangle XYZ is reflected across a coordinate grid what happens to the length and variables?"

Answer: It stays the same length and the variables turn into X' Y' Z'.

With this evidence its safe to say the correct choice is B.

Hope this answer helped you! I scrounge around Brainly to find people in my wake and assist them, don't be shy to give me Brainliest answer!

-Snooky

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(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

I am having a conflict with my family what is 7-1×0+3÷3 = ?

ryzh [129]

Follow PEMDAS.

Parenthesis
Exponents
Multiplication & Division
Addition & Subtraction

Alright, do multiplication first, then division. Then, do subtraction then addition.

$7-1*0+3/3= \\ 7-0+3/3= \\ 7 - 0 + 1= \\ 7 + 1 = 8 \\ 7-1*0+3/3 =8$

Hope this helped!

6 0

3 years ago

What is the measure of the indicated angle? 180° 6° 84° 96°

sleet_krkn [62]

Answer:

84°

Step-by-step explanation:

since the opposite side looks identical to the first angle, the second would be the same degrees.

3 0

2 years ago

Read 2 more answers

Need asapppp !!!!!!!

alina1380 [7]

X=70, y=55

The triangle is an isosceles triangles. In an isosceles triangle, the base angles are congruent.

Then use the triangle angle sum theorem and set the equation to 180.

55+55+x=180
110+x=180
x=70

The angle with 55 degrees and the angle with the variable y are alternate angles. Meaning they are congruent.

5 0

3 years ago

BRAINLIESTTT ASAP! PLEASE HELP ME :)

GrogVix [38]

Answer:

1. Type of function: absolute value function

2. The three transformations from the parent function f(x) = |x| are:

Translation 3 units to the left
Vertical stretch with a scale factor of 2
Translation 5 units downward

Explanation:

1. Type of function

The parent function |x| is the absolute value function. It returns the positive value of the argument (x).

Thus the function f(x) = 2|x + 3| - 5 is also an absolute value function.

It returns the positive value of x + 3, then multiplyes it by 2, and finally subtract 5.

This is a piecewise function.

For the values of x ≥ - 3, the output is 2(x +3) - 5 = 2x + 6 - 5 = 2x + 1.

For the values of x < - 3, the output is 2 (-x - 3) - 5 = -2x - 6 - 5 = -2x - 11.

The vertex of this function is at x = - 3: 2 (- 3 + 3) - 5 = 2(0) - 5 = 0 - 5 = 5.

Thus the vertex is (-3, 5).

2. Transformations from the parent function f(x) = |x|.

When you know the parent function and the daughter function you can know the transformations done of the former to get the later by some simple rules.

a) Translation in the horizontal direction.

When you add a positive value to the argument the function is translated to the left.

Thus, when you add 3 to x, |x| becomes |x + 3| and it is a translation 3 units to the left.

b) Vertical stretch

When you multiply the argument by a constant, the function stretches vertically.

Thus, when you multiply |x + 3| by 2, to get 2|x + 3|, the function is vertically stretched by a scale factor of 2.

c) Vertical translation

When you subtract a constant value from the function, you translate it downward.

Thus, when you subtract 5 from 2|x + 3| - 5, you translate the function 5 units downward.

6 0

3 years ago