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3 years ago

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Presley is marking the lines for a soccer field. She needs to form a right triangle near the goal. She measures the lengths of h

er sides as 9 yards, 6 yards and √45 yards. Did Presley form a right triangle? Why? Yes, because 6 squared + 9 squared = (startRoot 45 EndRoot) squared Yes, because 6 squared + (StartRoot 45 EndRoot) squared = 9 squared No, because 6 squared + 9 squared not-equals (StartRoot 45 EndRoot) Squared No, because 6 squared + (StartRoot 45 EndRoot) squared not-equals 9 squared

2 answers:

Hunter-Best [27]3 years ago

8 0

Answer:it’s b

Step-by-step explanation:

inessss [21]3 years ago

7 0

Answer:

B

Step-by-step explanation:

Yes, because 6 squared + (StartRoot 45 EndRoot) squared = 9 squared

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Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

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Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

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3 years ago