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2 years ago

What what is the instantaneous velocity of the caterpillar at time t=6

1 answer:

6 0

This problem is in Khan Academy

In this graph the instantaneous velocity=decline.
You just have to do this:

And the answer is letter (A)

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Find the circumference of this figure. Show your calculations.

solong [7]

Answer:

Circumference of circle= 37.68 units

Circumference of sector= 6.28 units

Step-by-step explanation:

The given figure is a circle with given radius OB= 6 units.

The circumference of a circle is the total length of its boundary.

For the complete circle its circumference can be calculated by;

Circumference= $2\pi r$

$=2*3.14*6\\=37.68$ units

For the circumference of the Sector with 60° in the figure, the circumference can be calculate by:

$=\frac{60}{360}* 2*\pi*6\\\\=\frac{1}{6}*12*3.14\\\\ =2*3.14\\\\=6.28$ units

So, the circumference of the full circle is 37.68 units and the circumference of the sector is 6.28 units.

5 0

2 years ago

Help me me me me plz thx if you do

prohojiy [21]

Answer:

75% left is not taken up so do the Ab to the other

Step-by-step explanation:

0.75 sorry if Im wrong

6 0

2 years ago

A city grid of Anytown, USA is shown on the grid below. The fire department is represented by quadrilateral RSTU. Another fire d

Alekssandra [29.7K]

Answer: option d. C (0,3), D (0,5).

Justification:

1) The x - coordinates of the vertices A and B are shown in the diagrama, They are both - 4, so the new vertices C and D must be in a line parallel to y = - 4.

2) The y-coordinates of the vertices A and B are also shown in the diagrama. They are equal to 3 and 5 respectively.

3) We can see that the new points C and D must be over a parallel line to y = - 4 and that their distance to the points A and B has to be the same distance of the point R and S to U and T.

That distance is 4, so the line may be y = - 7 or y = 0.

4) If the line is y = 7 the points C and D would have coordinates (-7,3) and (-7,5), but this points are not among the options.

5) If the line is y = 0 the points C and D would have coordinates (0, 3) and (0,5), which is precisely the points of the option d. That is the answer.

3 0

2 years ago

Read 2 more answers

Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.

fenix001 [56]

Answer:

The angle between vector $\vec{u} = 5\, \vec{i} - 8\, \vec{j}$ and $\vec{v} = 5\, \vec{i} + \, \vec{j}$ is approximately $1.21$ radians, which is equivalent to approximately $69.3^\circ$ .

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

their dot products, and
the product of their lengths.

To be precise, if $\theta$ denotes the angle between $\vec{u}$ and $\vec{v}$ (assume that $0^\circ \le \theta < 180^\circ$ or equivalently $0 \le \theta < \pi$ ,) then:

$\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ .

<h3>Dot product of the two vectors</h3>

The first component of $\vec{u}$ is $5$ and the first component of $\vec{v}$ is also

The second component of $\vec{u}$ is $(-8)$ while the second component of $\vec{v}$ is $1$ . The product of these two second components is $(-8) \times 1= (-8)$ .

The dot product of $\vec{u}$ and $\vec{v}$ will thus be:

$\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}$ .

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both $\vec{u}$ and $\vec{v}$ :

$\| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}$ .
$\| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}$ .

<h3>Angle between the two vectors</h3>

Let $\theta$ represent the angle between $\vec{u}$ and $\vec{v}$ . Apply the formula $\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}$ to find the cosine of this angle:

$\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}$ .

Since $\theta$ is the angle between two vectors, its value should be between $0\; \rm radians$ and $\pi \; \rm radians$ ( $0^\circ$ and $180^\circ$ .) That is: $0 \le \theta < \pi$ and $0^\circ \le \theta < 180^\circ$ . Apply the arccosine function (the inverse of the cosine function) to find the value of $\theta$ :

$\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ$ .

3 0

2 years ago

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

2 years ago