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MrMuchimi
3 years ago
12

Can someone please help me with the problems that aren't done please ASAP

Mathematics
2 answers:
Norma-Jean [14]3 years ago
4 0
Number 2. You first need to conver those two fractions to decimals to so 3/5 is 0.60 and 1/4 is 0.25 so this is the order: 1/4 , 0.4 , 0.55 ,3/5
Zanzabum3 years ago
3 0
I believe #8 would be 0.6, and #10 would be 65%.
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How many triangles???
Sergio [31]

Answer:

4

Step-by-step explanation:

The triangle itself

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6 0
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What is 6/2(1+2)=?
Rufina [12.5K]

Answer:

Answer is 9

Step-by-step explanation:

Remember PEMDAS???

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you would first solve the addition inside of the parentheses (1 + 2 = 3), and from there finish the equation as it's written from left to right.

8 0
2 years ago
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2(x + 3) = x - 4 <br>And <br>4(5x-2)=2(9x+3)<br>​
weqwewe [10]

2x+6 - х -4

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4 0
3 years ago
Part a at what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a di
loris [4]
3.01 m/s  
This is a simple projectile calculation. What we want is a vertical velocity such that the time the droplet spends going up and going back down to the surface exactly matches the time the droplet takes to travel horizontally 0.800 meters. The time the droplet spends in the air will is:
 V*sqrt(3)/2 ; Vertical velocity.
 (V*sqrt(3)/2)/9.8 ; Time until droplet reaches maximum height
 (V*sqrt(3))/9.8 ; Double that time for droplet to fall back to the surface.
 
 The droplet's horizontal velocity will be:
 V/2.
 
 So the total distance the droplet travels will be:
 d = (V*sqrt(3))/9.8 * V/2
 d = V^2*sqrt(3)/19.6
 
 Let's substitute the desired distance and solve for V
 d = V^2*sqrt(3)/19.6
 0.8 = V^2*sqrt(3)/19.6
 15.68 = V^2*sqrt(3)
 15.68/sqrt(3) = V^2
 15.68/1.732050808 = V^2
 3.008795809 = V 
 So after rounding to 3 significant figures, the archerfish needs to spit the water at a velocity of 3.01 m/s 
 Let's verify that answer.
 Vertical velocity: 3.01 * sin(60) = 3.01 * 0.866025404 = 2.606736465
 Time of flight = 2.606736465 * 2 / 9.8 = 0.531987034 seconds. 
 Horizontal velocity: 3.01 * cos(60) = 3.01 * 0.5 =
8 0
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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