For the seagull to catch the crab, h(t)=g(t) so:
-16t^2+45=-13t+23 add 16t^2 to both sides
45=16t^2-13t+23 subtract 45 from both sides
16t^2-13t-22=0 using the quadratic equation:
t=(13±√1577)/32, since t>0
t=(13+√1577)/32 seconds
t≈1.65 seconds (to nearest hundredth of a second)
And the height that this occurs using either original equation is:
h((13+√1577)/32)≈1.59 ft (to nearest hundredth of a foot)
Answer:
The correct option is 4.
Step-by-step explanation:
Let the line AB divided into 7 equal parts.
It is given that point P partitions the directed line segment from A to B in 3:4. It means 3 parts are before P and 4 partes are after P.
It is given that point Q partitions the directed line segment from A to B in 4:3. It means 4 parts are before Q and 3 partes are after Q.
It is given that point R partitions the directed line segment from A to B in2:5. It means 2 parts are before R and 5 partes are after R.
It is given that point S partitions the directed line segment from A to B in 5:2. It means 5 parts are before S and 2 partes are after S.
From the below figure we can say that point S is closest to point B.
It is also written as
Therefore option 4 is correct.
Tough sorry wish I could help u can’t too hard but I think it is equivalent to the other triangle so I think it’s 31? I THINK
Answer:
Step-by-step explanation:
surface area of one can 2πrh+2πr²
=2πr(h+r)
≈2×3.14×3/2(4.25+1.5)
≈9.42(5.75)
≈54.165 in²
number of cans=1000/54.165
=18 cans
material used=18×54.165≈975 in²
material left=1000-975=25 in²
Answer:
Claim 2
Step-by-step explanation:
The Inscribed Angle Theorem* tells you ...
... ∠RPQ = 1/2·∠ROQ
The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.
... 2·∠RPQ = ∠ROQ
The symmetric property of equality says you can rearrange this to ...
... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ
_____
* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.
You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.
