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3 years ago

PLZ HELP

Mathematics

1 answer:

anastassius [24]3 years ago

8 0

Answer:

A' (2,-3)

B' (5,-5)

C' (5,-2)

D' (7,-3)

Step-by-step explanation:

I used desmos graphing calculator to help me solve

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Please answer this question now in two minutes

Igoryamba

Answer: x=6, y=20

Step-by-step explanation:

Since ΔBCD and ΔTUS are congruent triangles, we can set the sides equal to each other.

y=2y-20 [subtract both sides by 2y]

-y=-20 [divide both sides by -1]

y=20

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3x+32=9x-4 [add both sides by 4, and subtract both sides by 3x]

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3 years ago

Alice drives to Alabama at a rate of 60pmh. She drives home at a rate of 40mph. If the total trip takes her 5 hours, how long do

jonny [76]

OK, so she definitely drove home faster than when she drove to her first destination. So she must've taken quicker to get home. Lets assume 3 hours on the way there and 2 hours on the way back.
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3 years ago

Name an inscribed angle

Vladimir79 [104]

Answer:

BHF

Step-by-step explanation:

Definition of inscribed

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3 years ago

Nathan goes to the store and buys 2 shorts and 4 shirts for 70$. He later goes to the same store and buys 3 shorts and 3 shirts

stiks02 [169]

Answer:

Each item of clothing costs $12.50

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Step-by-step explanation:

8 0

3 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago