Answer:
it takes 2 seconds to do one situp
it takes 2 seconds to do one pushup
to do 14 pushups, it takes 28 seconds.
to do 12 situps it takes 24 seconds
to do 12 pushups, it takes 24 seconds
BRAINLIEST PLEASE!
hope this helps :)
Step-by-step explanation:
14x + 12y = 52
12x + 12y = 48
^^^ this is an inequality that we can make.
x stands for pushups
y stands for situps
if we subtract the second equation from the first, we get
2x = 4
x = 2
now we can place x in one of the equations to solve for y
12(2)+12y=48
24 + 12y = 48
12y = 24
y = 2
Answer:
(-1/4, 19/4)
Step-by-step explanation:
5x+6=9x+7
-4x=1
x= -1/4
y=5(-1/4) +6
y=19/4
There are two ways you can do this. You can use the longer way:
15+16+17+18+19+20+21+22+23+24 = 195 (starting from the top row, which has 15 logs, up until the bottom row, which has 24 logs, on condition that each second row has one log more than the previous one)
You can also use the formula (where n is the number of rows)
x= (n(first term+last term))/2
x= (10(15+24))/2
x= (10*39)/2
x=390/2
x=195
Either way, there are 195 logs in the stack.
Answer:
f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)
Step-by-step explanation:
f(x) = eˣ
Expand using first Taylor Polynomial based around b = 0
The Taylor's expansion based around any point b, is given by the infinite series
f(x) = f(b) + xf'(b) + (x²/2!)f"(b) + (x³/3!)f'''(b) + ....= Σ (xⁿfⁿ(b)/n!) (Summation from n = 0 to n = ∞)
Note: f'(x) = (df/dx)
So, expanding f(x) = eˣ based at b=0
f'(x) = eˣ
f"(x) = eˣ
fⁿ(x) = eˣ
And e⁰ = 1
f(x) = 1 + x + (x²/2!) + (x³/3!) + ....... = Σ (xⁿ/n!) (Summation from n = 0 to n = ∞)
Answer:
Step-by-step explanation:
