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2 years ago

Write the function for the graph.

Mathematics

1 answer:

artcher [175]2 years ago

3 0

Answer:

$f(x) =3 * (4)^x$

Step-by-step explanation:

Given

$(x_1, y_1) = (1,12)$

$(x_2, y_2) = (0,3)$

Required

The function of the graph (exponential function)

An exponential function is represented as:

$f(x) = ab^x$

For $(x_1, y_1) = (1,12)$ , we have:

$12 = a * b^1$

$12 = a * b$

$12 = a b$

For $(x_2, y_2) = (0,3)$

$3 = a * b^0$

$3 = a * 1$

$3 = a$

$a = 3$

Substitute: $a = 3$ in $12 = a b$

$12 = 3 * b$

Divide both sides by 3

$4 = b$

$b =4$

So, we have:

$f(x) = ab^x$

$f(x) =3 * (4)^x$

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Walking at a constant rate of 8 kilometers per hour, juan can cross a bridge in 6 minutes. What is the length of the bridge in m

Mama L [17]

Answer:

800 m

Step-by-step explanation:

1 km = 1,000 m

8 km = 8,000 m

1 hour = 60 minutes

6 minutes is 1/10 of an hour

1/10 of 8,000 m = 800 m

Answer: 800 m

7 0

3 years ago

Ft?

VARVARA [1.3K]

Answer:

15 5/8 ft^3

Step-by-step explanation:

Volume of a cube = l x b x h

Given the Length of sides is

2 1/2

Change 2 1/2 to an improper fraction.

It becomes 5/2

Volume = 5/2 x 5/2 x 5/2

Multiply through

= 125/8

= 15 5/8 ft^3

5 0

3 years ago

Find the midpoint between (-1+9i) and B=(5-3i)

sdas [7]

Answer:

2 + 3i, midpoint is (2,3)

Step-by-step explanation:

we need to find the midpoint between (-1+9i) and B=(5-3i)

To find the midpoint of two points (a+bi) and (c+di) in a complex plane,

we apply formula

$\frac{a+c}{2} + \frac{b+d}{2} i$

A = (-1+9i) and B=(5-3i)

Midpoint for AB is

$\frac{-1+5}{2} + \frac{9+(-3)}{2}i$

$\frac{4}{2} + \frac{6}{2}i$

2 + 3i , so midpoint is (2,3)

8 0

3 years ago

Read 2 more answers

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

3 years ago

I don't get this I really need help to understand this "/

Alisiya [41]

I don't either it is quite hard to be fair

4 0

3 years ago