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4 years ago

(1/2)^3+(8*3/4)-6 please help asap

Mathematics

1 answer:

Natali [406]4 years ago

7 0

(1/2)^3 = 1/2 * 1/2 * 1/2 = 1/8

8 * (3/4) = (8 * 3)/4 = 24/4 = 6

Then, from order of operations, you can do the subtraction first or the addition first. The subtraction would be easier to do first, because we would have 6 - 6 = 0.

Then, we’re left with 1/8 + 0 which is just 1/8.

The answer is 1/8.

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What is the greatest common factor of the terms in the polynomial 8x4 – 4x3 – 18x2?

RSB [31]

Answer:

2x^2

Step-by-step explanation:8 divides by 2

4 divides by 2 and 18 divides by 2

It can not by 4 because 18/4 is not a whole number it's a decimal

8 0

4 years ago

Questions are in the pictures

Elan Coil [88]

The values of h and r to maximize the volume are r = 4 and h = 2

<h3>The formula for h in terms of r</h3>

From the question, we have the following equation

2r + 2h = 12

Divide through by 2

r + h = 6

Subtract r from both sides of the equation

h = 6 - r

Hence, the formula for h in terms of r is h = 6 - r

<h3>Formulate a function V(r)</h3>

The volume of a cylinder is

V = πr²h

Substitute h = 6 - r in the above equation

V = πr²(6 - r)

Hence, the function V(r) is V = πr²(6 - r)

<h3>The single critical point</h3>

V = πr²(6 - r)

Expand

V = 6πr² - πr³

Integrate

V' = 12πr - 3πr²

Set to 0

12πr - 3πr² = 0

Divide through by 3π

4r - r² = 0

Factor out r

r(4 - r) = 0

Divide through by 4

4 - r = 0

Solve for r

r = 4

Hence, the single critical point on the interval [0. 6] is r = 4

<h3>Prove that the critical point is a global maximum</h3>

We have:

V = πr²(6 - r)

and

V' = 12πr - 3πr²

Determine the second derivative

V'' = 12π - 6πr

Set r = 4

V'' = 12π - 6π* 4

Evaluate the product

V'' = 12π - 24π

Evaluate the difference

V'' = -12π

Because V'' is negative, then the single critical point is a global maximum

<h3>The values of h and r to maximize the volume</h3>

We have

r = 4 and h = 6 - r

Substitute r = 4 in h = 6 - r

h = 6 - 4

Evaluate

h = 2

Hence, the values of h and r to maximize the volume are r = 4 and h = 2

Read more about maximizing volumes at:

brainly.com/question/1869299

#SPJ1

7 0

2 years ago

If f(x)=16x-30 and g(x)=14x-6, for which value of x does (f-g)(x)=0

tamaranim1 [39]

Answer:

x=12

Step-by-step explanation:

f(x)=16x-30 and g(x)=14x-6

(f-g)(x)=0

f(x)=16x-30 -(14x-6)

Distribute

= 16x -30 -14x +6

Combine like terms

= 16x-14x -30+6

2x-24

Set this equal to zero

2x-24 =0

Add 24 to each side

2x-24 +24=0+24

2x=24

Divide by 2

2x/2 =24/2

x = 12

5 0

3 years ago

Please help!!!

viktelen [127]

Answer:

eight and a half a night for 6th graders and they are 23 countries in North America

Step-by-step explanation:

idk what statistical means sorry

7 0

3 years ago

Read 2 more answers

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago