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2 years ago

Show that the equation x3 + 4x = 1 has a solution between x = 0 and x = 1

1 answer:

5 0

Intermediate Value Theorem: Suppose that f(x) is an arbitrary, continuous function on an interval [a,b] . If there exists a value L between f(a) and f(b) , then there exists a corresponding value c∈(a,b) , such that f(c)=L

f(x)=x3+4x−1

f(0)=−1f(1)=4

Since the function changes sign in the interval (0,1) , hence there exists a c∈(0,1) such that f(c)=0

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You are designing an open-top cylindrical container. The cylinder must have a volume of 81π cm3 . The bottom of the container mu

stira [4]

Answer:

Minimum dimensions are r=3cm, h=9cm

Minimum Cost=$254.47

Step-by-step explanation:

Volume of a Cylinder=πr²h

Volume of the Open Top Cylinder=81π cm³.

Therefore:

πr²h=81π

The bottom costs $3 per cm² and the side costs $1 per cm².

Total Surface Area of the open top Cylinder= πr²+2πrh

Cost, C=3πr²+2πrh

As the Volume is fixed.

πr²h=81π

r²h=81

h=81/r²

Modifying C,

$C=3\pi r^{2}+2 \pi r \frac{81}{r^{2}}$

$C=3\pi r^{2}+ \frac{162 \pi}{r}$

We differentiate C with respect to r

C'= $6\pi r -\frac{162 \pi}{r^2}$

At the minimum cost, C'=0.

Next we solve C'=0 for r

$6\pi r -\frac{162 \pi}{r^2}=0$

6πr³-162π=0

6πr³=162π

r³=27

r=3

The dimensions of the cylinder at minimum cost are therefore:

r=3 cm

h=81/9=9cm

The minimum cost of the Cylinder

C=3πr²+2πrh

=(3XπX3²)+(2XπX3X9)

=$254.47

8 0

4 years ago

Read 2 more answers

Which expression is equivalent to

KonstantinChe [14]

Given the expression:

$\sqrt[4]{ \frac{24x^{6}y}{128x^{4}y^{5}}}$

Simplifying:

$\sqrt[4]{ \frac{3x^{2}}{16y^{4}}}$

Then the answer is:

$\frac{ \sqrt[4]{3x^{2}}}{2y}$

6 0

3 years ago

The graph of a polynomial function of degree 5 has three x-intercepts, all with multiplicity 1. Describe the nature and number o

mr Goodwill [35]

For this case suppose that we have a polynomial in its standard form of the form:

$f (x) = ax ^ 5 + bx ^ 4 + cx ^ 3 + dx ^ 2 + ex + f$

Where,

a, b, c, d, e, f: coefficients of the polynomial

x: independent varible

f (x): dependent variable

Since the polynomial is of degree 5, then the polynomial has 5 solutions that comply with:

$f (x) = 0$

We know that three of the solutions are real and are repeated only once.

Therefore, the two remaining solutions are complex numbers.

Answer:

C) The function has 3 real and 2 imaginary zeros.

6 0

3 years ago

Read 2 more answers

How do you solve this 14g>56

melomori [17]

Answer:

g >4

Step-by-step explanation:

14g>56

Divide each side by 14

14g/14 > 56/14

g >4

6 0

3 years ago

U ONLY HAVE TO SHOW ME WHERE TO PLACE THEM NOT THE SECOND THING ONLY THE FIRST ONE BUT IF U DO THE SECOND ONE TO I WILL GIVE U B

DIA [1.3K]

Answer:

for the 10 am gor to the number 1 and go up 6 little lines

for the 12:00 place it on 0

for 6:30 go to -5 and go down 2 lines

Step-by-step explanation: hope this help

5 0

3 years ago