<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%7B2%7D%5E%7Bx%7D%20%20%20%2B%203" id="TexFormula1" title="f(x) = {2}^{x} +

All categories

Anon25 [30]

3 years ago

3" alt="f(x) = {2}^{x} + 3" align="absmiddle" class="latex-formula">
What is the value of f(-2)

Mathematics

1 answer:

lisov135 [29]3 years ago

4 0

The function is given $f(x)=2^x+3$ .

Now let's solve for $f(-2)$ .

$f(-2)=2^{-2}+3$

Rule of negative exponent: $x^{-1}=\frac{1}{x^1}$ .

$f(-2)=\frac{1}{2^2}+3$

$f(-2)=\frac{1}{4}+\frac{3}{1}$

Rule for sum of fractions: $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ .

$f(-2)=\frac{13}{4}$

And the result is:

$f(-2)=\boxed{\frac{13}{4}=3.25}$

You might be interested in

Three straight lines have exactly two points of intersection complete the sentence below : two of the lines must be .....

olga nikolaevna [1]

Paralell i forgot how to spell it lol

4 0

3 years ago

The time, t, required to drive to a certain distance varies inversely with the speed, r. if it takes 12 hours to drive the dista

kap26 [50]

T = k/R where k is the constant of variation, T=time and R=speed.
Plug in the given values:-
12 = k/60
k = 12*60 = 720
So the equation of variation is T = 720/R

At 85 mph we have

T = 720 / 85 = 8.47 hours (answer)

8 0

3 years ago

Please help me! thank you so much

Luda [366]

The Correct Answer Is f(x)=2.5x+5

7 0

3 years ago

Explain how you could use a number line to show that -4+3 and 3+(-4) have the same value.which property of addition states that

Anna007 [38]

For -4+3,
1. start with zero
2. move to left (-x axis) for 4 units (-4)
3. then move to right (+x axis) for 3 units (+3)

For 3+(-4)
1. Start with zero
2. Move the point to right side by 3 units (+3)
3. then move the point to the left by 4 units (-4)

For both you'll stop at -1.

8 0

3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$