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3 years ago

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What is equivalent to (1/4 + y)^4

2 answers:

Mrac [35]3 years ago

7 0

Hope it will help you!!

LuckyWell [14K]3 years ago

6 0

You can expand this using the binomial theorem

= 1/256 + 4C1 (1/4)^3 y + 4C2 (1/4)^2 y^2 + 4C3 (1/4) y^3 + y^4

= 1/256 + 1/16 y + 3/8 y^2 + y^3 + y^4

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<img src="https://tex.z-dn.net/?f=5%5E%7B-3%7D" id="TexFormula1" title="5^{-3}" alt="5^{-3}" align="absmiddle" class="latex-form

AveGali [126]

Answer:

the equivalent equation of

$5^{-3}$ = $\frac{1}{5^{3}}=\frac{1}{125}$

5 0

3 years ago

Blake is a songwriter who collects royalties on his songs whenever they are played in a commercial or a movie. Blake will earn $

Mrac [35]

Answer:

It was played in 3 movies and in 10 commercials.

Step-by-step explanation:

110 times 3 = 330

530 minus 330 = 200

200 divided by 20 = 10

3 0

3 years ago

Darrel divided 575 by 14 by using partial quotients, what is the quotient?

Usimov [2.4K]

143r3 (143 remainder 3)

3 0

3 years ago

A manufacturer has been selling 1000 flat-screen TVs a week at $500 each. A market survey indicates that for A manufacturer has

luda_lava [24]

Answer:

a) Demand function: $q=6000-10\cdot p$

b) The rebate should be of $200, so the sale price becomes $300 per unit.

c) The rebate should be of $150, so the sale price becomes $350 per unit.

Step-by-step explanation:

a) In this case, we have a known point of the demand function (1000 units sold at $500), and the slope of a linear function (increase by 100 units fora decrease in $10).

We can express the demand function (linear) as:

$q=b+m\cdot p$

To calculate the slope m, we use:

$m=\Delta q/\Delta p=(+100)/(-10)=-10$

To calculate b, we use the known point and the calculated slope:

$q=b+m\cdot p\\\\b=q-m\cdot p=1000-(-10)\cdot (500)=1000+5000=6000$

Then the demand function is:

$q=6000-10\cdot p$

b) The revenue can be expressed as:

$R=q\cdot p = (6000-10p)\cdot p=6000p-10p^2$

To maximize, we can derive and equal to zero

$dR/dp=6000-2*10p=0\\\\20p=6000\\\\p=300$

The rebate should be of $200, so the sale price becomes $300 per unit.

c) If we take into account the cost, we have that

$R=q\cdot p-C=(6000p-10p^2)-(68000+100q)\\\\R=(6000p-10p^2)-(68000+100(6000-10p))\\\\R=6000p-10p^2-(68000+600000-1000p)\\\\R=-10p^2+7000p-668000$

To maximize, we can derive and equal to zero

$dR/dp=-20p+7000=0\\\\p=7000/20=350$

The rebate should be of $150, so the sale price becomes $350 per unit.

5 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago