Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
Frequency
Amplitude
Wavelength
Speed
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
Answer:
f = 55mm, h ’= -9.89 cm
f = 200 mm, h ’= 42.5 cm
Explanation:
For this exercise let's start by finding the distance to the image, using the equation of the constructor

where f is the focal length, p and q are the distances to the object and image, respectively
lens with f₁ = 55mm = 0.55cm
=
= 1.718
q₁ = 0.582 m
lens with f₂ = 200mm = 2m
=
= 0.4
q₂ = 2.5 m
the magnification of a lens is given by
m =
h ’=
let's calculate for each lens
f = 55mm
h '= - 0.582 / 10 1.7
h ’= 0.0989 m
h ’= -9.89 cm
f = 200 mm
h '= - 2.5 / 10 1.7
h ’= -0.425 m
h ’= 42.5 cm
The negative sign indicates that the image is real and inverted
Answer:
<h3>C. </h3>
Explanation:
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