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algol13
3 years ago
5

Please help asap! Thank youu

Mathematics
1 answer:
noname [10]3 years ago
7 0

if it's likely but not certain i would think it's c and if it's wrong i'm very sorry

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Help me please! Thank you.
Alex_Xolod [135]

f'(x) = 9+1/x

f'(9) = 9+1/9 = 82/9


6 0
4 years ago
Solve for b:4 = -2/3(5) + b
saveliy_v [14]

Answer:

\frac{22}{3} = b

or

7\frac{1}{3} = b

(they are the same value, just different ways of writing it)

Step-by-step explanation:

4 = -\frac{2}{3} (5) + b

simplify

4 = -\frac{10}{3} + b

now use inverse operations

4 = -\frac{10}{3} + b

+\frac{10}{3}  +\frac{10}{3}

\frac{22}{3} = b

4 0
4 years ago
If f(x) = x+5 then fill the table <br>x| 1 | | 2 | |<br>y| | 3 | |5 |​
notsponge [240]

Answer:

6,-2,7,0

Step-by-step explanation:

Use the equation and plug in values

f(1)=1+5

y=6

--

3=x+5

x=-2

--

f(2)=2+5

y=7

--

5=x+5

x=0

4 0
3 years ago
t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.
kondaur [170]

Answer:

a) t = 1, b)  t = -2, c) Concave upward: (-\infty, 1). No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}

By factorizing each component, the function is re-arranged as:

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

There is no avoidable discontinuities. The only point where tangent line is vertical is t = 1.

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is  horizontal is t = -2.

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}

f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]

f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}

f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}

The only critical point is:

-6\cdot (t+2) = 0

t = -2 (which coincides with the result of point b)

f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]

f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}}  \right]

The value associated with the critical point is:

f''(-2) = \frac{2}{9}

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is (-\infty, 1). There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0
4 years ago
........... Help........​
FrozenT [24]

jill will make less money if they work for 4 hours

6 0
4 years ago
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