<img src="https://tex.z-dn.net/?f=16y%20%5E%7B5%7D%20-%2012y%20%5E%7B4%7D%20%2B%204y%20%5C%5C%20%5C%5C%2064%20a%20%5Ctimes%20%5E

All categories

1 year ago

%7B2%7D%20-%2049ay%20%5E%7B2%7D%20" id="TexFormula1" title="16y ^{5} - 12y ^{4} + 4y \\ \\ 64 a \times ^{2} - 49ay ^{2} " alt="16y ^{5} - 12y ^{4} + 4y \\ \\ 64 a \times ^{2} - 49ay ^{2} " align="absmiddle" class="latex-formula">
Plss answer this

Mathematics

1 answer:

Shkiper50 [21]1 year ago

7 0

Answer:

see explanation

Step-by-step explanation:

Assuming you require to factorise the expressions

16 $y^{5}$ - 12 $y^{4}$ + 4y ← factor out 4y from each term

= 4y(4 $y^{4}$ - 3y³ + 1)

--------------------------------

64ax² - 49ay² ← factor out a from each term

= a(64x² - 49y²) ← this is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

64x² - 49y²

= (8x)² - (7y)²

= (8x - 7y)(8x + 7y)

then

64ax² - 49ay² = a(8x - 7y)(8x + 7y)

You might be interested in

Find the value of x (7x - 12) 114

Harrizon [31]

Answer:

(7x-12)=-5

-5x÷-5 114÷-5=

x=29

5 0

2 years ago

Read 2 more answers

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito th

Likurg_2 [28]

Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$ , while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

$=e^{\int \:\left(\frac{1}{50}\right)dt}$

$=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

$=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) Plot the Solution to see the ultimate behavior of the graph

The graph appears to level off at about the value of $Q=25$ .

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$ .

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

$A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

Learn more about differential equation word problem from brainly.com/question/14614696

#learnwithBrainly

4 0

3 years ago

How much candy at $1.16 a pound should be mixed with candy worth 86 cent a pound in order to obtain a mixture of 60 pounds of ca

vovangra [49]

Let the weights of the two candies be repres. by x and y.

Then x + y = 60, or x = 60 - y

($1.16 / lb) x + ($0.86 / lb) y = ($1.00 / lb) (60 lb) = $60

Then 1.16(60-y) + 0.86y = 60
69.6 - 1.16y + 0.86y = 60 9.6
9.6 = 0.3y Solving for y, y = ------- = 32 lb
0.3

Then x = (60-32) lb = 28 lb

5 0

3 years ago

Read 2 more answers

The owner of a quick oil-change business charges

Mkey [24]

Answer:

the sun is actually a planet

Step-by-step explanation:

8 0

3 years ago

Y=Cosx.arcsinx What is the solution

Alinara [238K]