6x9x(-5) what is the answer to this question?

Need help ASAP! (Geometry 1)

melisa1 [442]

Answer:

The Length of third side is 52 units.

Step-by-step explanation:

Given:

Length of First side = 48

Length of second side = 20

Also given the angle between them is 90°.

Hence we can say figure shown is a right angled triangle.

For finding the third side we can use Pythagoras theorem which states that;

"If a triangle is right angled then, the Square of the hypotenuse is equal to sum of the square of the other two sides."

$Third \ side ^2=First \ side^2+Second \ side^2$

Hence substituting the values we get;

$Third\ side^2= 48^2+20^2 = 2304+400 =2704$

Now taking Square root on both side we get;

$\sqrt{Third \ side^2} = \sqrt{2704}\\ Third \ side = 52 \ units$

Hence Length of Third side is 52 units.

3 0

3 years ago

are there any other considerations that should be taken into account when trying to fit a giant tv into your car

FrozenT [24]

The width and height of the tv as well as the length and width of the back of your car

6 0

2 years ago

Bla nnn,,,,,,bs jfbjnf msnjkmfe sjnmf cmmnefnmv fcve

EleoNora [17]

Answer:

Ummmmm ehhhhhh

8 0

2 years ago

Read 2 more answers

Conduction can occur within a single substance, but parts of the substance must have which property? different temperatures diff

Julli [10]

Answer:

A,

Step-by-step explanation:

answer is A

6 0

2 years ago

Read 2 more answers

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago