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)
First one: If the leading coeff. is negative, the graph begins in Quadrant III and ends in Quadrant IV. It's an even function. The fourth graph represents it.
If the leading coeff. is positive, but everything else remains the same, the graph opens upward, beginning in Q II and ending in Q I.
Answer:
Step-by-step explanation:
Sin 21 =
Sin 21 =
0.3583 =
0.3583*15 = BC
BC = 5.3745 = 5.37 m
Tan ∅ =
Tan ∅ =
∅ = tan⁻¹ (0.7448) = 36.67 = 36.7
Step-by-step explanation:
To solve this, you either need x or y. So x or y can be the missing values. The most you can do is simply move one of the terms to the other side.
The answer is 8x I believe