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4 years ago

The difference between longitude and altitude

1 answer:

5 0

Hello there,

The difference are Longitude (shown as a vertical line) is the angular distance, in degrees, minutes, and seconds, of a point east or west of the Prime (Greenwich) Meridian. Lines of longitude are often referred to as meridians.

And altitude is the height of an object or point in relation to sea level or ground level , great height

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Your friend Ana received an 90% on a quiz. If Ana got 36 problems right, how many questions were in the quiz?

kherson [118]

Answer:

Step-by-step explanation: I just guesstimated a number and divided 36 by 40 to get .9 which means 90%

3 0

3 years ago

3. Find the measure of x. a 18° b. 54° C 126 d. 45

Nataliya [291]

Answer:

18 degrees

Step-by-step explanation:

The triangle is an iscoceles right triangle.

The angles in a triangle add up to 180.

90+2y (iscoceles) =180

2y=90

y=45

So the angles of the right triangle are 45. However, you have to take away 27 because you are solving for only a part of 45. 45-27=18

6 0

3 years ago

Read 2 more answers

Ms. Kork sold her car for $8700. This was $100 more than two fifths of what she had paid for the car originally. How much mad Ms

Ne4ueva [31]

Answer:

8600 is 100 less

3440?

Step-by-step explanation:

3 0

3 years ago

hi people i suck at math and is in 10th grade i have a question if you start at 5,2 and move 3 units left and 2 units up what do

irina [24]

Answer:

(2,4)

Step-by-step explanation:

left( - ) and right( + ) (x axes)

up( + ) and down( - ) (y axes)

(5,2) just add

5 - 3 = 2

2 + 2 = 4

Answer :

(2,4)

8 0

3 years ago

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

6 0

3 years ago