PLEASE HELP ME FAST QUESTION IN PHOTO WILL PICK BRAINLIEST IF CORRECT
1 answer:
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The sum of two vectors is (- 0.5, 10.1)
<u>Explanation: </u>
To add two vectors, add the corresponding components.
Let u =⟨u1,u2⟩ and v =⟨v1,v2⟩ be two vectors.
Then, the sum of u and v is the vector
u +v =⟨u1+v1, u2+v2⟩
(b)
Two vectors = ( 3, 4 )
angle 2π/3 = 120°
In x axis, the vector is = 7 cos 120°
In y axis, the vector is = 7 sin 120°
= 7 X 0.866
= 6.062
The second vectors are ( -3.5, 6.062)
Sum of two vectors = [( 3 + (-3.5) ), (4 + 6.062)]
= (- 0.5, 10.1)
<h3>Answer: x = 7 and y = 3</h3>
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Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
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Let's consider case A
We have this system of equations
Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
----------------
Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
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Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.
Answer:
B) x units
Step-by-step explanation:
Let quadrilateral KMPT be a rectangle with dimensions 12 units by 8 units. Then its perimeter would be equal to:
Perimeter of a rectangle = 2 (l + b)
where: l is the length = 12 units and b is the breadth = 8 units. So that:
Perimeter of KMPT = 2 (12 + 8)
= 40 units
Dilating KMPT by a scale factor of would create K'M'P'T' of dimensions;
× 12 units by
× 8 units. Thus, the dimensions of K'M'P'T' would be 9 units by 6 units.
Perimeter of K'M'P'T' = 2 (l + b)
= 2(9 + 6)
= 30 units
Comparing the perimeters of KMPT and K'M'P'T', the perimeter of K'M'P'T' would be × perimeter of KMPT.
Therefore, if the perimeter of KMPT is x units, then;
perimeter of K'M'P'T' = * x units
= x units