Can someone please help me please
1 answer:
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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?
Answer:
Step-by-step explanation:
Given:
A stuntman jumping off a 20-m-high building is modeled by the equation
-----------(1)
A high-speed camera is ready to making film between 15 m and 10 m above the ground
when the stuntman is 15m above the ground.
height
Put height value in equation 1
We know that the time is always positive, therefore
when the stuntman is 10m above the ground.
height
Put height value in equation 1
Therefore ,time interval of camera film him is
Answer:
CD ≈ 26.0 cm
Step-by-step explanation:
using the sine ratio in right triangle ABD
sin35° = =
=
( multiply both sides by BD )
BD × sin35° = 12 ( divide both sides by sin35° )
BD = ≈ 20.92 cm
using the sine rule in Δ BCD
=
, that is
=
( cross- multiply )
CD × sin52° = 20.92 × sin102° ( divide both sides by sin52° )
CD = ≈ 26.0 cm ( to 3 significant figures )
Answer:
Solution given:
1:
-1/8=
2. -64/27
=
Express in rational number
1. (-3/2)=-3/2*2/2=-(3*2)*(2/2)=-6/4
2. (1/5)=1/5*5/5=<u>5/25</u>
and
(-4/3)³(2/5)-⁴ ÷ (7/4)
(-4³/3³)(2-⁴/5-⁴)÷7/4
(-64/27)(5⁴/2⁴)÷7/4
(-64/27)(625/16)÷7/4
(-64*625/(27*16))*4/7
-2500/27*4/7
-10000/169
(-100/13)²