Answer:
2 4/16 = 2 1/4 lb
Step-by-step explanation:
The pointer is 4 of the smallest units above 2. There are 16 of those small units. Each 2 of those units is 1/8 pound, so when the pointer is on a multiple of 2 units, it is on a multiple of 1/8 pound. No guesswork is required to choose the correct weight to the nearest 1/8 pound.
2 4/16 = 2 1/4 lb . . . . the scale reading to the nearest 1/8 pound.
Answer:
f'(x) > 0 on 
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:
To find its decreasing interval :
2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72
3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
Answer:
Distance traveled, initial speed, and final sped.
Step-by-step explanation:
A)Plugging in our initial statement values of y = 16 when x = 10, we get:
16 = 10k
Divide each side by 10 to solve for k:
16/10=
k = 1.6
Solve the second part of the variation equation:
Because we have found our relationship constant k = 1.6, we form our new variation equation:
y = 1.6x
Since we were given that x, we have
y = 1.6()
y = 0
B)Plugging in our initial statement values of y = 1 when x = 15, we get:
1 = 15k
Divide each side by 15 to solve for k:
1/15
=15k
k = 0.066666666666667
1.
a) metres to centimetres :
multiply length by 100
b) metres to millimetres:
multiply length by 1000
c) kilograms to grams:
multiply the mass value by 1000
d) litres to millilitres :
multiply volume by 1000
2.
a) 3 m = 3× 100 = 300 cm
b) 28 cm = 28 × 10 = 280 mm
c) 2.4 km = 2.4 × 1000
= 24 × 10^-1 × 10^3
= 24 × 10^2 =2400 m
d) 485 mm =485 / 10
= 485 / 10 ^1
= 485 × 10 ^-1
= 48.5 cm
e) 35 cm = 35 / 100
= 35 /10^2
= 35 × 10 ^ -2
= 0.35 m
f) 2.4 m = 2.4 / 1000
= 24 × 10 ^-1 / 10^3
= 24 × 10^-1 × 10 ^-3
= 24 × 10 ^ -4
= 0.0024 km
g) 2495 mm = 2495 /1000
= 2495 /10^ 3
= 2495 × 10 ^-3
=2.495 m
