If you would like to know what is the constant for the quadratic equation 9 * y = 4 * x^2, you can calculate this using the following steps:
<span>9 * y = 4 * x^2 /9
</span>y = 4 * x^2 / 9
y = 4/9 * x^2
The correct result would be B. 4/9.
Answer:
(i) x° = 28°
(ii) y° = 104°
(iii) z° = 76°
Step-by-step explanation:
x° is an alternate interior angle where transversal PQ crosses the parallel lines, so it has the same measure as the one marked 28°.
x° = 28°
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The base angles of isosceles triangle PQR are both z°, so we must have ...
z° +z° +28° = 180° . . . . . . . . . sum of angles in a triangle
z° = (180° -28°)/2 = 152°/2 . . . solve for z
z° = 76°
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y° and z° are a linear pair, so ...
y° = 180° -76°
y° = 104°
Answer:
first of all, the graphs are so poorly plotted,even kn computer...
anyways,
since the blue one ending at x=5 on the x axis,
while actual log ends at x=0 this means
X=x-5
and since, log1=0 but it's 5 in blue graph,
the graph is moved up by 5
hence,
y=5+log(x-5)
When 36 is decreased by 75%, the result is 9.
75% of 36 is 27.
36 - 27 = 9
PART 1.
Going westward:
Distance: 2400 miles
Rate (this is speed): (450-wind_speed) mph
Time: distance / speed = 2400 / (450-wind speed)
Going eastward:
Distance: 2400 miles
Rate (this is speed): (450+wind_speed) mph
Time: distance / speed = 2400 / (450+wind speed)
PART 2
Formula for speed:
speed = distance / time
Now we solve this for time:
time = distance / speed
11h = 2400 / (450+-wind_speed)
+- sign is put because in one case wind increases speed of plane and in other case it decreases speed.
PART 3
To solve for wind_speed we start by rearranging formula so that we have wind_speed on left side:
(450+-wind_speed) = 2400 / 11
(450+-wind_speed) = 218mph
To fly this distance in 11h speed of plane must be 218mph.
The only way to get this solution is if we substract wind speed:
450-wind_speed = 218
wind_speed = 450-218 = 232mph
