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4 years ago

Four times y to the fourth power

Mathematics

2 answers:

alexandr1967 [171]4 years ago

8 0

(x 2 y) 4 = x2*4 y 1*4 = x 8 y4

cluponka [151]4 years ago

5 0

$4{ y }^{ 4 }\\ \\ ={ y }^{ 4 }+{ y }^{ 4 }+{ y }^{ 4 }+{ y }^{ 4 }\\ \\ =4\left( y\cdot y\cdot y\cdot y \right)$

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Evaluate function f(x)= x. For where x=3−π

alexira [117]

Answer:

if the shown formula is correct/complete,

f(3−π) would equal just 3−π, or about -0.14159

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3 years ago

Given: S,T, and U are Collinear and ST = TÚ

slava [35]

It's true
collinear means they are on the same line so ST=SU
so if ST= 50 then SU=50 making T the midpoint of line that is 100
s----------------t-----------------u
50 50

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3 years ago

Simplify the expression (10 + 3i)(10 − 3i).

Stels [109]

Answer:

Step-by-step explanation:

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4 years ago

Read 2 more answers

Drag the correct steps into order to evaluate 16 + 20t for t = 4.

densk [106]

Answer:

see below

Step-by-step explanation:

16+20/t

Let t=4

16 + 20/4

First we divide

16 + 5

Then we add

8 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago