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LekaFEV [45]
3 years ago
14

Which functions have a range of.. SELECT MORE THAN 1

Mathematics
1 answer:
liraira [26]3 years ago
7 0

Answer:

The 3rd and the 5th

Step-by-step explanation:

The graphs of the 3rd and the 5th functions are straight lines

The 1st and 2nd are parabolas

The 4th is an exponential

You must look at the graph: it shall include all the real values for the y coordinate

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Solve for x. Round to the nearest tenth, if necessary.<br> B<br> X<br> A<br> 7.9<br> 54°
Alja [10]

Answer:

I got 10.87 , which rounded would be 10.9

Step-by-step explanation:

because you use tan, opposite/adjacent

so it is 7.9tan54 to get your answer :)

x = 10.873417172

3 0
3 years ago
Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(
BlackZzzverrR [31]
We can parameterize this part of a cone by

\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle

with 0\le u\le a and 0\le v\le2\pi. Then

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv

The area of this surface (call it \mathcal S) is then

\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi
6 0
3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLEIST!!
Finger [1]

i think it is 1.5 percent just divide

8 0
3 years ago
What Is The Slope-Intercept Form Of This Equation 3x+9y=18
alexgriva [62]
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6 0
3 years ago
Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont
Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

f(x) = x^{3}-9\cdot x^{2}+ 3, 2 \leq x \leq 12

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

f'(x) = 3\cdot x^{2}-18\cdot x

3\cdot x^{2}-18\cdot x = 0

3\cdot x \cdot (x-6) = 0 (Eq. 1)

As we can see, only a solution is a valid critical value. That is: x = 6

Second, we determine the second derivative formula and evaluate it at the only critical point:

f''(x) = 6\cdot x -18 (Eq. 2)

x = 6

f''(6) = 6\cdot (6)-18

f''(6) =18 (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

f(2) = 2^{3}-9\cdot (2)^{2}+3

f(2) = -25

x = 6

f(6) = 6^{3}-9\cdot (6)^{2}+3

f(6) = -105

x = 12

f(12) = 12^{3}-9\cdot (12)^{2}+3

f(12) = 435

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

6 0
3 years ago
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