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3 years ago

Help me, please I could fail

Mathematics

1 answer:

balu736 [363]3 years ago

5 0

Answer: I believe the answer is B 8/25.12

Step-by-step explanation:

Tell me if this helps

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Help please gbnhhbbnhfgnrjttnettyjnry

mafiozo [28]

The third one is the answer

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3 years ago

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The midpoint of the line segment from P1 to P2 is (-6,5). If P1=(-9,7), what is P2?

PolarNik [594]

Let point O be the midpoint of a segment P₁P₂. Then point O has coordinates (-6,5). You also know coordinates (-9,7) of point P₁.

Use formula for midpoint's coordinates:

$x_O=\dfrac{x_{P_1}+x_{P_2}}{2} \text{ and } y_O=\dfrac{y_{P_1}+y_{P_2}}{2}.$

Substituting known coordinates, you get:

$-6=\dfrac{-9+x_{P_2}}{2} \text{ and } 5=\dfrac{7+y_{P_2}}{2}.$

Thus,

$x_{P_2}=-6\cdot 2+9=-12+9=-3,\\ \\y_{P_2}=5\cdot 2-7=10-7=3.$

Answer: $P_2(-3,3).$

6 0

3 years ago

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A figure with 4 sides and angles. totalling 360°

Arlecino [84]

Answer:

a acute square

Step-by-step explanation:

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4 years ago

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Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$