Rewrite the following polynomial in standard form. 1−2x−5x^4 +8x^3

What is the weight of an object whose mass is 16.4 kg?

MrRissso [65]

The answer is 161.70 N.

EXPLANATION

To change from mass to weight multiply by gravity (9.8 m/s^2)

Fw = m * 9.8 m/s^2
Fw = 16.5 kg * 9.8 m/s^2 = 161.70 N.

4 0

3 years ago

Read 2 more answers

84 out of 240 what is the percentage

Nesterboy [21]

Convert fraction (ratio) 84 / 240 Answer: 35%

5 0

2 years ago

Read 2 more answers

Mr. Moretti is making welcome bags to give to the 18 students in his homeroom on the first day of school he puts m mechanical pe

Mashcka [7]

Answer:

The total number of pencils is 6

Step-by-step explanation:

According to the given scenario, the computation of the total number of pencils is as follows:

There are 18 students

And let us assume there is m mechanical pencils that put in each bag

also he put twice of the regular pencil

So if he put the 3 mechanical pencil

So, the total number of pencils is

= 3 × 2

= 6

As it twice of the regular pencil

hence, the total number of pencils is 6

5 0

2 years ago

Givenf(x)=6(8−x)+5. (a) What does the notation f (−2) mean? (b) What is the value of f (−2) ?

Nonamiya [84]

Answer:

b. 65

a. The notation f(-2) means that we want to evaluate the value of the equation when x = -2

Step-by-step explanation:

A. The notation f(-2) means that we want to evaluate the value of the equation when x = -2

B. We will simply substitute the value of 2 for x

f(x) = 6(8-x) + 5

f(-2) = 6(8 - (-2)) + 5

f(-2) = 6(8 + 2) + 5

f(-2) = 6(10) + 5 = 60 + 5 = 65

5 0

2 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

2 years ago