Answer:
14%
Step-by-step explanation:
28/200 = x/100
x = 14
1.
Solution here,
let the both planes covers same distance x while over taking.
For the small plane, let time be t.
speed= 240 mph
now,
distance covered by it,
x=240t--------(1)
For jet plane,let the time be t'.
speed=36mph
since it is flewed after 30 mins, it can be written as,
t'=t-30 min=t-0.5 hr
now distance covered by it,
c=360t'=360(t-0.5)=360t-180----------(2)
equating (1) and (2)
240t=360t-180
or, -120t=-180
or, t=1.5 hr
therefore two planes wiil meet after 1.5 hrs.
putting the value of t in (1)
x=240×1.5=360 m
therefore they travel through 360 m from the airport whlie over taking.
2.
For the first car,
speed=50 Kmph
let it covers x distance at time t
so,diatance x=50t--------(1)
For the second car,
speed=45 Kmph
according to the question, in time t, it will covers the distance (x-20)Km
so, distance(x-20)=45t
or, x=45t+20---------(2)
equating (1) and (2),
50t=45t+20
or, 5t=20
or, t=4 hrs.
therefore cars will be apart of 20 km after 4 hrs.
Hi there!
Since the sum of all angles in any triangle always adds up to 180 degrees, we can calculate this and solve for x.
4x + x + x = 180°
6x = 180
180 ÷ 6 = x
x = 30
So, now we know that x is 30°.
Hope this helps!
Answer:
The answer is
Step-by-step explanation:
f(x) & g(x) = f(x) + k
3•6 & 7•6 = 3•6 + k
18 + 42 = 18k
60 = 18k
\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤
