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Cerrena [4.2K]
1 year ago
7

V8 to the nearest tenth is about ?

Mathematics
1 answer:
777dan777 [17]1 year ago
6 0
\sqrt[]{8}\approx2.8

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What is the definition of marginal​ utility? A. The average utility from consuming a good or service multiplied by the number of
Ilya [14]

Answer:  Option 'E' is correct.

Step-by-step explanation:

Marginal utility is the addition utility derived from each successive unit of a particular good.

so, marginal utility is the change in utility from consuming an addition unit of a good or service.

Mathematically, it is expressed as

MU=\dfrac{\Delta TU}{\Delta Q}

Hence, Option 'E' is correct.

4 0
3 years ago
Add the following complex numbers: (5-3i)+(10+5i)
serious [3.7K]

Answer:

i = 7.5

Step-by-step explanation:

(5-3i)+(10+5i)

(5+10)+(5i-3i)

15+2i

15=-2i

15/2-i

7.5=i

6 0
3 years ago
Find the measure of the indicated angle to the nearest degree.
makvit [3.9K]
You would do tan<span>Ɵ = 9/12
tan^-1(9/12) = </span><span>Ɵ
</span><span>Ɵ = 36.87 degress
make sure your calculator is in degree mode

</span>
8 0
3 years ago
Read 2 more answers
What is the equation of the circle with center (16, - 12) and a radius of 5
eimsori [14]

Answer:

x^2+y^2-32x+24y+375=0

Step-by-step explanation:

Equation of circle with given centre (h,k) and radius=r is

(x-h)^2+(y-k)^2=r^2

Given centre are=(16, - 12) and radius= 5

So equation of circle=

(x-16)^2+(y- -12)^2=5^2

x^2-32x+256+y^2+24y+144=25

x^2+y^2-32x+24y+375=0

Answer

5 0
3 years ago
Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

as desired. So one possible ODE would be

-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

(See "Euler-Cauchy equation" for more info)
6 0
3 years ago
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