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2 years ago

On the plane curve defined by the parametric equations below, what point corresponds to t= -1

2 answers:

6 0

The equation x=2t-1 represent the x-coordinate. To get the value of x, substitute t with -1.

That is: x = 2(-1) - 1
= -2-1
= -3
Likewise, the equation y=t∧-3 represent the y-coordinate. Substituting t with -1,

y = (-1)∧-3
= -1

The point that correspond to t=-1 is (-3,-1).

AlladinOne [14]2 years ago

6 0

Answer: I just took the test and got 100%

1. C, the parameter is t and the curve contains the set of points (2cos(t), sin(t)

2. B, up to the left and then up to the right

3. D, (-3,-1)

Feel free to mark as brainliest (:

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3 years ago

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base

Olenka [21]

Answer:

The base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = $x\times\frac{1}{2} =\frac{x}{2} \ m.$

The height of triangle = $x\times \frac{3}{4} =\frac{3x}{4}\ m.$

The area of triangle = $16\ m^2.$

Now, we put the formula of area to solve:

$Area=\frac{1}{2} \times base\times height$

$16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}$

$16=\frac{3x^2}{16}$

Multiplying both sides by 16 we get:

$256=3x^2$

Dividing both sides by 3 we get:

$\frac{256}{3} =x^2$

Using square root on both sides we get:

$\frac{16}{\sqrt{3}}=x$

$x=\frac{16}{\sqrt{3}}$

Now, by substituting the value of $x$ to get the base and height:

$Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.$

So, the base of triangle = $\frac{8}{\sqrt{3}} \ m.$

$Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.$

Thus, the height of triangle = $\frac{12}{\sqrt{3}} \ m.$

Therefore, the base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

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