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4 years ago

What is U(2.4, -1) reflected across the y-axis?

Mathematics

2 answers:

AVprozaik [17]4 years ago

8 0

Answer:

(-2.4, -1)

Step-by-step explanation:

When you mirror a point across the y-axis, which is the vertical axis, the x-coordinate changes it's sign from either positive to negative or negative to positive.

stich3 [128]4 years ago

6 0

Answer:

(-2.4,-1)

Step-by-step explanation:

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The high temperatures on April 1 in Jackson, Austin Carson City,and Des Moines were 73F,42,F,and 40F respectively. Write the nam

kondor19780726 [428]

Des Moines - 40F
Austin Carson City - 42F
Jackson - 73F

3 0

3 years ago

How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

Line AB

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

Line BC

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

Line AC

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

Perimeter of Triangle ABC

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

What is the value of X? 30 60 36 48

Alisiya [41]

Take 36 and add it to 60 it gives you 96 then divide it by 2 and the answer is 48

3 0

3 years ago

Read 2 more answers

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$