So

x=hours
y=pay

use a piecewise function maybe

for x≤40, use y=21x

for x>40, use 840+1.5(21(x-40))

k, sso 46>40 so

840+1.5(21(46-40))
840+1.5(21(6))
840+1.5(126)
840+189
1029

A. 840+1.5(21(x-40))
B. $1029

7 0

3 years ago

If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo

Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or, Q = 40 kg × 4200 × 18

or, Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0

3 years ago

What is the axis of symmetry for the function f(x)=3(x-2)^2+4

Alex73 [517]

y=3(x-2)²+4

Vertex is(2,4)

Axis of symmetry: x=2

5 0

3 years ago