we know that
The measure of the interior angle is the half-summit of the arcs that comprise it and its opposite
so
therefore
the answer is
the measure of arc TU is equal to 
Let daniel's age = d and let jessica's age = j
let your current age = x and your wife's current age = y
x = 5d (given in question)
∴ x - 5d = 0
x + 21 = 2(d+21) (given in question)
∴ x + 21 = 2d + 42
∴ x - 2d = 21
We can solve these two rearranged equations simultaneously by multiplying the first equation by -1 and adding them. This gives us the following:
3d = 21
∴ d = 7
This means that daniel is currently 7 and (if we substitute d = 7 into one of the equations) you are 35. We use a similar method for your wife's and jessica's current ages.
y = 7j (given in question)
∴ y - 7j = 0
y + 8 = 3(j + 8)
∴y + 8 = 3j + 24
∴ y - 3j = 16
If we use a similar method of elimination we get this:
4j = 16
∴ j = 4
Hence, from this we can concur that daniel is 7 and jessica is 4.
Answer:
Step-by-step explanation:
speed =65m/mins = 65/60 = 1.083m/s
Time taken to go down,
t = 4485/1.083
= 4141.37 s
Total time gotten = 12 hrs = 12*60*60
= 43200s
Amount of time left to conduct experiment
= 43200-4141.37
= 39058.63s
= 39058.63/3600
= 10.85 hrs
∠ M ≅ ∠ R: true
<span>VL ≅ LT: true
</span><span>Δ MLV can be rotated about point L to map it to Δ RLT. : false
</span><span>A series of rigid transformations of Δ MLV maps it to Δ RLT. : true </span>
Answer: (6-9) Hatchlings would be a reasonable outcome for the simulation.
As the histogram shows the number of chicks born to non-migratory Canada geese in a city park, with the horizontal axis representing the number of hatchlings and the vertical axis representing the number of nests.
In order to monitor the geese population, the state wildlife service annually samples the number of hatchlings from 30 nests using a simulation= 6-9 hatchlings.
(As there is only 1 bar with (6-9)hatchlings which is meeting to the 30 nests on y axis)
