3) Altitude / Time = y2 - y1 / x2 - x1 = 30 - 60 / 6 - 3
m = -30 / 3
m = -10
In short, constant rate of change is y = -10x
b) Constant proportionality exists between two quantities, as the amount of changing in Altitude over fixed period of time is same (constant) for every instance.
4) Sales / Day = y2-y1 / x2-x1 = 2,000 - 1,000 / 6 - 3
m = 1000 / 3
m = 333.3
a) In short, Constant relationship is y = 333.3x
b) Constant proportionality exists between two quantities, as the amount of changing in Sales over fixed days is same (constant) for every instance.
Hope this helps!
So the ratio to longboard to short board is 1:4, the total of that ratio is 5. 35 ÷ 5 = 7. now we're going to take those ratios and multiply 1 × 7 = 7 and 4 × 7 = 28. Your final answer is B. to confirm your answer add 28 + 7.
Qn. 1
Lower bound for Zoe's weight = 62 - (1/2) = 62 - 0.5 = 61.5 kg
Qn. 2
Upper bound for length AB = 8.3+ (0.1/2) = 8.3+0.05 = 8.35 cm
Qn. 3
Upper bound for Anu's wight = 83+(0.5/2) = 83+0.25 = 83.25 kg
Qn. 4
Lower bound for length CD = 27-(0.5/2) = 27-0.25 = 26.75 cm
Qn. 5
Upper bound for sides of the hexagon = 3.6+(0.1/2) = 3.6+0.05 = 3.65 cm
Upper bound for the perimeter = upper bound for the sides*6 = 3.65*6 = 21.9 cm
Qn. 6
Perimeter = 4*length => side = Perimeter/4 = 24/4 = 6
Bound = 0.5/4 = 0.125
Lower bound of the length = 6-0.125 = 5.875 cm
Qn. 7
For the area,
Upper bound = 80+(10/2) 80+5 = 85 cm^2
For the length
Upper bound = 12+(1/2) = 12+0.5 = 12.5
Then, upper bound for the width = Upper bound for the area/upper bound for the length = 85/12.5 = 6.8 cm
Qn. 8
Lower bound for the area = 230-(1/2) = 230-0.5 = 229.5 cm^2
Lower bound for the sides of the square = Sqrt(Lower bound of the area) = Sqrt (229.5) = 15.15
Then,
Lower bound of perimeter = 4(Length) = 4*15.15 = 60.6 cm
Answer:
![X = \left[\begin{array}{cccc}12&-24&8&16\end{array}\right]](https://tex.z-dn.net/?f=X%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D12%26-24%268%2616%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
A matrix is given by
![A = \left[\begin{array}{cccc}6&-12&4&8\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D6%26-12%264%268%5Cend%7Barray%7D%5Cright%5D)
Now, we have to find the matrix X if 
⇒ X = 2A
Now, multiplication of a constant with a matrix means multiplication with that constant with all the terms of the matrix.
So, ![X = \left[\begin{array}{cccc}6 \times 2&-12 \times 2&4 \times 2&8 \times 2\end{array}\right]](https://tex.z-dn.net/?f=X%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D6%20%5Ctimes%202%26-12%20%5Ctimes%202%264%20%5Ctimes%202%268%20%5Ctimes%202%5Cend%7Barray%7D%5Cright%5D)
⇒ (Answer)
Answer:
20,24,28
Step-by-step explanation:
It is increasing by 4 each time
Mark me brainliest please :)
