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3 years ago

Write the equation in exponential form using the two points (0,1) and (1,3)

Mathematics

1 answer:

MakcuM [25]3 years ago

8 0

Answer:

$y=3^x$

Step-by-step explanation:

The equation of the function in exponential form is

$y=a\cdot b^x$

The function is determined using points (0,1) and (1,3), so their coordinates satisfy the eduation. Substitute them:

$1=a\cdot b^0\Rightarrow a=1\ \ [b^0=1]\\ \\3=a\cdot b^1\Rightarrow 1\cdot b=3,\ b=3$

Thus, the equation of the function is

$y=3^x$

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Let t = 4 and u = 6 + 2i. Find t – u

hjlf

Step-by-step explanation:

t = 4

u = 6 + 2

Find t - u

solu

t - u

= 4 - ( 6 + 2 ) = 4 - 8

Answer is -4

5 0

2 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$