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3 years ago

Which is the graph of f(x) = 0.5(4)x?

Mathematics

2 answers:

Sloan [31]3 years ago

6 0

Answer:

The first graph, so A is the answer

Step-by-step explanation:

Natali5045456 [20]3 years ago

5 0

Find f(x) for x = 0

f(x) = 0.5(4^x)

f(0) = 0.5 * 4^0

f(0) = 0.5 * 1 = 0.5

The graph must contain point (0, 0.5).

Now find f(1).

f(x) = 0.5(4^x)

f(1) = 0.5 * 4^1

f(1) = 0.5 * 4 = 2

The graph must contain point (1, 2).

Now find f(2)

f(x) = 0.5(4^x)

f(2) = 0.5 * 4^2

f(2) = 0.5 * 16 = 8

The graph must contain point (2, 8).

Now find f(3)

f(x) = 0.5(4^x)

f(3) = 0.5 * 4^3

f(3) = 0.5 * 64 = 32

The graph must contain point (3, 32).

Find a graph that has the points (0, 0.5), (1, 2), (2, 8), and (3, 32).

The only graph that contains those 4 points is choice A.

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Find the distance between X(-3, 8) and Z(-1,5) *<br> what is it :’)

ahrayia [7]

Answer:

d = √13

Step-by-step explanation:

d = √((x2 - x1)² + (y2 - y1)²)

points: (-3, 8) & (-1, 5)

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d = √(2)²+(-3)²

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3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

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3 years ago

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ch4aika [34]

Answer:

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Step-by-step explanation:

each exponent needs to multiply each 10.

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3 years ago

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Step-by-step explanation:

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3 years ago

A car travels 2 5/8 miles in 3 1/2 minutes at a constant speed. which equation represents the distance that the car travels in m

Ksju [112]

Answer:

3/4 mile per minute

Step-by-step explanation:

2 5/8=21/8

3 1/2=7/2

---------------

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3 years ago