HIII Brainly People I need help with those math problems I HATE MATH so if you could help that me that would be nice. :)

Rus_ich [418]

Answer:

15771 > - 7

Step-by-step explanation:

Comparing the elevation of the two points ;

Point with elevation of 15771 is positive meaning it occurs above sea level and this higher than the point with an elevation of - 7 which is occurring below sea level

Hence, to compare both elevations :

Montblanc has higher elevation than Rhone river delta

Montblanc elevation > Rhone river delta

15771 > - 7

4 0

3 years ago

Which of the following correctly completes the square for the equation below?

erastova [34]

The answer is :

c. (x+1)²=20

To check:
(x+1)²=20 (a+b)²=a²+2ab+b²
x²+2x+1=20
x²+2x=20-1
x²+2x=19

3 0

3 years ago

Read 2 more answers

Using the image, find the distance between the points given on the graph.

Yuliya22 [10]

Answer:

3 units

Step-by-step explanation:

using the distance formula

6 0

3 years ago

Read 2 more answers

What is the domain and range of the function f(x)=−3/2(4)^(x−3)−1?

Mekhanik [1.2K]

Answer:

Domain: (−∞,∞)

Range: (−∞,∞)

5 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 $\theta$ :

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

3 0

3 years ago