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timofeeve [1]
3 years ago
5

27. (17xyz) Taking the square root of perfect squares

Mathematics
1 answer:
sveticcg [70]3 years ago
4 0

Step-by-step explanation:

download bitbox helps with these kinda questions

and hopefully u find what your looking for in there make sure to brainlist please ! nd contact anytime

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The perimeter of parallelogram ABCD is 46 inches.What is DA?
Shkiper50 [21]

Answer:

3

Step-by-step explanation:

4 0
3 years ago
Exploring Volume of Pyramids
S_A_V [24]

Answer:2h

Step-by-step explanation:

7 0
3 years ago
Which of the following equations represents the area of a sector?
Rina8888 [55]

From the given options we can say that the only one that represents the area of the sector is; A = n/360 * πr²

<h3>What is the Area of the Sector?</h3>

In circles, a sector is said to be a part of a circle made of the arc of the circle together with its two radii. This means that it is a portion of the circle formed by a portion of the circumference (arc) and radii of the circle at both endpoints of the arc.

The formula for Area of a sector is given as;

θ/360 * πr²

where;

θ is the central angle of the sector

r is radius

Now, looking at the given options we can say that the only one that represents the area of the sector is;

A = n/360 * πr²

where n is the central angle of the sector

Read more about Area of Sector at; brainly.com/question/16736105

#SPJ1

4 0
1 year ago
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
3 years ago
Please help! I'm getting timed.
Molodets [167]
Hi there! The answer is A.

t = 2(s + 2)
To get a proper idea of the type of function this formula represents we work out the parenthesis.

Working out the parenthesis can for instance be done using rainbow technique.
t = 2s + 4

We can now see sinilarities between this function an the general function of a line in slope intercept form
y = mx + n

The function t therefore describes a line with slope 2 and y-intercept 4. Therefore the answer is A.
3 0
3 years ago
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