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mihalych1998 [28]
3 years ago
11

In general, the point

Mathematics
1 answer:
Zanzabum3 years ago
8 0

Answer:

we conclude that when we put the ordered pair (0, a), both sides of the function equation becomes the same.

Therefore, the point (0, a) is on the graph of the function f(x) = abˣ

Hence, option (D) is correct.

Step-by-step explanation:

Given the function

f(x) = abˣ

Let us substitute all the points one by one

FOR (b, 0)

y = abˣ

putting x = b, y = 0

0 = abᵇ

FOR (a, b)

y = abˣ

putting x = a, y = b

b = abᵃ

FOR (0, 0)

y = abˣ

putting x = 0, y = 0

0 = ab⁰

0 = a         ∵b⁰ = 1

FOR (0, a)

y = abˣ

putting x = 0, y = a

a = ab⁰

a = a         ∵b⁰ = 1

TRUE

Thus, we conclude that when we put the ordered pair (0, a), both sides of the function equation becomes the same.

Therefore, the point (0, a) is on the graph of the function f(x) = abˣ

Hence, option (D) is correct.

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You throw a baseball into the air at an initial velocity of 30 feet per second. When the ball leaves your hand it is 5 feet abov
ValentinkaMS [17]
Assuming this little game of catch took place on planet earth, negative acceleration due to gravity is -9.8 m/s .
Converting 30ft/s to m/s, initial velocity was 30ft/s x 0.305ft/meter = 9.14m/s 

Let's find how long it took for the velocity to equal zero, meaning when the ball reached it's highest point and, for a split second, stopped in mid-air before falling back down.

V(t) = Vi + a*t , where V(t) is velocity as a function of time, a is acceleration due to gravity, and t is time. Set V(t) = 0
0 = 9.14 + (-9.8)* t          Add -9.8t to both sides
9.8t = 9.14                      Divide both sides by 9.8
t = 0.93 seconds

Let's say your hand is the base point, or where h=zero. We want to find how high above your hand the ball went before it started coming down. Using the distance, or in this case height, formula:
h = Vi*t + (1/2)at²               Plug in Vi, a, and our t value, 0.93
h= 9.14 * 0.93 + (1/2)(9.8)(0.93²)
h= 8.5 + 4.9 (0.865)
h = 8.5+ 4.27
h = 12.74 meters

The ball made it 12.74 meters above your hand. Your friends had was one foot above yours, so let's subtract .305 meters to see how far it dropped from the peak height to his hand.
12.74-.305 = 12.43 meters

Let's use the distance formula again to see how long it took to come down. Remember that this time, initial velocity is zero, since the ball starts off suspended in the air.
-12.43 = 0*t + (1/2)(-9.8)(t²)         Divide both sides by -9.8/2, or -4.9
2.5374 = t²
t = 1.59

The ball took .93 seconds to go up, and 1.59 seconds to come down to your friend's glove. The total time the ball was in the air:
.93 + 1.59 = 2.52 seconds
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A. 20 x 6
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