Answer:
I=\frac{t^2}{2}
Step-by-step explanation:
From exercise we have that x=t, t>0. Because A(t) be the area of the region in the first quadrant, we get that x started at 0. The limits for y are the following e-x and e. We get the integral :
I=\int\limits^0_t \int\limits^{e}_{e-x} 1 dy dx
I=\int\limits^0_t [y]_{e-x}^{e} dx
I=\int\limits^0_t (e-e+x) dx
I=\int\limits^0_t {x} \, dx
I=[\frac{x^2}{2} ]_{0}^{t}
I=\frac{t^2}{2}
Answer:
-15625
Step-by-step explanation:
Hope this helps!
Answer: 17.7.
59.4 * 0.28 = 59 * 0.3
Step-by-step explanation: ---------->
59 * 0.3 = ?
59 * 0.3 = 17.7
Your answer is 17.7.
I hope this helps!
3 × 100 + 9 × 10 + 1 + ( 6 × 1 ⁄10 ) + ( 7 × 1 ⁄100 ) + ( 5 × 1 ⁄1000 )
Answer:
$19.35
Step-by-step explanation:
129 x .15 = 19.35
