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4 years ago

Which matrix is equal to

Mathematics

1 answer:

polet [3.4K]4 years ago

3 0

The one that look exactly the same it the third one

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Help help help ASAP

kirill115 [55]

The answer is: x = 13

8 0

2 years ago

PLEASE HELPP WILL GIVE BRAINLIEST!!

umka2103 [35]

Well the line is straight so it will always have the same slope no matter what points you use but to solve it mathematically you do change in y/change in x

(3.5-(-1))/(4-(-5))
4.5/9 = 0.5
So the slope is 0.5

Now pick 2 random points on the line
(-3,0) and (5,4)
Now do the same thing for these 2 points

(4-0)/5-(-3)
4/8 = 0.5

So no matter what points the slope is always 0.5

3 0

3 years ago

Read 2 more answers

What is the shortest distance from point to a straight line is

Len [333]

So, if we define a straight line to be the one that a particle takes when no forces are on it, or better yet that an object with no forces on it takes the quickest, and henceshortest route between two points, then walla, the shortest distance between two points is the geodesic; in Euclidean space, a straight line as

3 0

4 years ago

I am sus lol so here is some points

Masja [62]

Hello.

Have a beautiful and joyful day ahead

Thanks for points..

3 0

3 years ago

Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the

s344n2d4d5 [400]

Answer:

80cm^2/s

Step-by-step explanation:

This is a related rates problem where we are considering the rate at which the area of a square changes with respect to time.

So lets consider the area of a square:

A = s^2 (where s represents the length of one side of the square)

Related rates problem deal with functions of time so if we take the area and side length as a function of time and then differentiate implicitly we get:

$\frac{dA}{dt} = 2s(\frac{ds}{dt})$

The problem states that the side of a square is increasing at a rate off 8cm/s so we can conclude that ds/dt = 8cm/s leaving us with:

$\frac{dA}{dt} = 16s$

Now, to solve for s we have to consider the other value given. If the area of the square is initially 25cm^2 we can plug this into our formula for area to solve for the side length.

25 = s^2

s = +/- 5 (since side lengths are only positive we only consider +5)

s = 5

Now we can plug this back in for s:

$\frac{dA}{dt} = 80$

Therefore, the rate at which the area of the square is increasing is 80cm^2 per second.

5 0

4 years ago