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lesya692 [45]
3 years ago
15

HELP ME QUICK!!! this has to be turned in in 46 min

Mathematics
1 answer:
IRISSAK [1]3 years ago
4 0

Answer:

the answer is most likely A

Step-by-step explanation:

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The length of a rectangle is 5in longer than its width. If the perimeter of the rectangle is 46in, find its area
Crank
<span>width = x
</span><span>length = x+5

2x + 2(x+5) = 46
2x + 2x + 10 = 46
4x = 46 - 10
4x = 36
x = 36/4 </span><span>
x = 9 in  </span> ←  width 

length = x+5 = 9 + 5 = 14 in
<span>
Area = 9 * 14 = 126 in</span>²<span>

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3 years ago
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Which is an asymptofe of the graph the function y=cot(x-2π/3)
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C) x=4π/3


The variable x in the cotangent argument has a unit coefficient, so the period is π, just as it is in the parent function cot(x).


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3 years ago
How much greater is the surface area of the rectangular prism than the surface area of the cube?
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3 years ago
Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona
andrey2020 [161]

Answer:

a. v.v' = v₁v₂ -  v₁v₂ = 0 b.  (20, -21)/29 and  (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' =  (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j  

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

=  v₁v₂ -  v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0
2 years ago
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