Finding the surface area

Mathematics

1 answer:

Serga [27]2 years ago

7 0

Answer: so you do length times width times height and then you divide by two

Step-by-step explanation:

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Type the correct answer in each box. Round the vector's magnitude to the nearest tenth.

Mama L [17]

Component form of u is (-18,13) and The magnitude of u is 22.2

<h3>What is a vector?</h3>

A vector is a two-dimensional entity with both magnitude and direction. A vector can be visualized geometrically as a directed line segment.

The component form of a vector is an ordered pair that describes the change in x and y values

This is mathematically expressed as (Δx, Δy) where Δx=x₂-x₁ and Δy=y₂-y₁

Given ;

Initial points of the vector as (14,-6)

The terminal point of the vector is (-4,7)

Here x₁=14,x₂=-4, y₁=-6,y₂=7

The component form of the vector u is (-4-14,7--6) =(-18,13)

Finding the Magnitude of the vector

u=√(x₂-x₁)²+(y₂-y₁)²

u=√-18²+13²

u=√324+169

u=√493

u=22.2

Therefore the Component form of u is (-18,13) and The magnitude of u is 22.2

To know more about vectors follow

brainly.com/question/25705666

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3 0

2 years ago

you are to give a client one tablet labeled 0.15 mg and one labeled 0.025 mg. what is the total dosage of these two tablets

Ivenika [448]

Simply add the two dosages together, (0.15+0.025) and the answer is 0.175 :)

7 0

3 years ago

How do u do b??????????

horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

$\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x$

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

$\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x$

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

$\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5$