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3 years ago

What is the range of the following number sequence 1,2,3,4,5,6,7,8,9

Mathematics

2 answers:

Lelechka [254]3 years ago

5 0

Answer:

Step-by-step explanation:

To find range, subtract the lowest number(1) from the highest number(9)

highest-lowest

9-1

So, the range is 8

Hope this helps! :)

FrozenT [24]3 years ago

3 0

Answer:

Step-by-step explanation:

To find the range, we put the values in order from smallest to largest

1,2,3,4,5,6,7,8,9

The we take the largest number and subtract the smallest number

9-1 =8

The range is 8

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Write the equation of the circle centered at ( 4 , − 5 ) with radius 18.

Free_Kalibri [48]

Answer:

(x-4)^2 + (y+5)^2 = 324

Step-by-step explanation:

The equation of a circle is given by the equation (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center point of the circle.

Therefore, since the center point of the circle and the radius is given, we can just plug the numbers into the formula:

(x-4)^2 + (y+5)^2 = 18^2

3 0

3 years ago

PLEASE SOME ONE HELP ME IM DYING!!!!!

Afina-wow [57]

We know the area of the middle rectangle is 48 (length * width). removing that rectangle leaves us with two semicircles. you can combine those semicircles to be the equivalent of one circle. the area for a circle is r^2 * pi. we know the diameter is 4 because that is where we cut the semicircles. radius is half the diameter, so r is 2. 2^2 is 4, 4* pi is 12.56. add 12.56 (area of semicircles) with 48 (area of rectangle) and we get 60.56

4 0

3 years ago

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 $\theta$ :