Most people are saying 5 and 1/3, but I think it's something else. Could be a poorly worded question or me not reading correctly or a bit of both. Fraction of 4 = 4 is the whole. is 3/4 = 3/4 is the part. You're trying to figure out the proportion of 4 that 3/4 takes up. In other words, (3/4)/4, which is 3/16. This is actually the opposite of 5 and 1/3. This could just be me reading the question wrong but that's how I interpret it. Answer from YoshiMachbike12 on reddit.com
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Reddit
reddit.com › r/math › simple math riddle: a proof that 4 = 3
r/math on Reddit: Simple math riddle: a proof that 4 = 3
June 1, 2012 -

Probably easier for some of you than it was for me, but I thought it was a fairly fun one:

Theorem: 3=4
Proof:

Suppose:

a + b = c  

This can also be written as:

4a - 3a + 4b - 3b = 4c - 3c

After reorganizing:

4a + 4b - 4c = 3a + 3b - 3c

Take the constants out of the brackets:

4 * (a+b-c) = 3 * (a+b-c)

Remove the same term left and right:

4 = 3
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Mathway
mathway.com › popular-problems › Basic Math › 74439
Solve for t (t-3)/2=(4-3t)/4 | Mathway
Solve for t (t-3)/2=(4-3t)/4 · Step 1 · Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction. Step 2 · Solve the equation for .
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Solve for t: 3/4 t=-4
The solution is t=-16/3
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symbolab.com
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3/4 t=-4
What is the answer to 3/4 t=-4 ?
The answer to 3/4 t=-4 is t=-16/3
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3/4 t=-4
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Symbolab
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3/4 t=-4
Trigonometry · Statistics · Physics · Chemistry · Finance · Economics · Conversions · Go · Hide Plot » · View interactive graph > 17=3(p-5)+8 · solve\:for\:x,x+3y=-1(1) x+\frac{3}{4}=4+2x · \frac{2}{3}x+5=3 · \frac{8+x}{2}-\frac{x+3}{3}=4 · The answer to 3/4 t=-4 is t=-16/3 ·
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Mathway
mathway.com › popular-problems › Precalculus › 432917
Solve for t t-4/t=3 | Mathway
If any individual factor on the left side of the equation is equal to , the entire expression will be equal to . Step 3.4 · Set equal to and solve for . Tap for more steps... Step 3.4.1 · Set equal to . Step 3.4.2 · Add to both sides of the equation. Step 3.5 · Set equal to and solve for .
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cs.stackexchange.com › questions › 104787 › solving-the-recurrence-tn-n3-41-4-n
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You can just expand it: $$ ... \cdot n^{3/16} \cdot n^{3/64} T(n^{1/64}) \\ &= \cdots \end{align*} $$ We see that $T(n) = C(n) n$, where $C(n)$ is the number of times we need to apply $n \mapsto n^{1/4}$ until the answer ...
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Answer (1 of 14): 2/3/4 = 2*4/3 = 8/3. The simple logic - when dividing by a number, multiply the dividend with the multiplicative inverse of the divisor - is used. As 2 is divided by 3/4, the calculation process involves multiplying 2 by the inverse of 3/4, which is 4/3. 2 multiplied by 4/3 g...
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What is the answer to 4/3? - Quora
4/3 means 4 divided by 3 but is usually called “four thirds” as though that is just a number. In some sense it is just a number, but in another sense it retains an arithmetic...
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Reddit
reddit.com › r/learnmath › [linear algebra] t:r4 to r4 satisfies t^3 + 3t^2 = 4i. s = t^4 +3t^2 -4i.
r/learnmath on Reddit: [Linear Algebra] T:R4 to R4 satisfies T^3 + 3T^2 = 4I. S = T^4 +3T^2 -4I.
May 4, 2023 -

Apologies if you saw the other post. I made a mistake in writing the equations down.

____________________________________________________________________________________________________________

T:R4 to R4 satisfies T3 + 3T2 = 4I. S = T4 +3T2 -4I.

S is :

one one but not onto.

onto but not one one

invertible

noninvertible

my answer : S is non invertible

____________________________________________________________________________________________________________

Applying Cayley Hamilton, eigenvalues of T is 1,-2,-2.

When we look at S, eigenvalue 1 -> 1+3-4 = 0,

S has a 0 eigenvalue. S is singular. Thus S cannot be one-one or onto or invertible.

But the answer key answer is one one but not onto..

When dimension of domain and codomain are the same, doesn't one one imply onto?

Another question is T is from dimension 4. But I only have 3. Does that imply one eigenvalue was 0 and was hastily cancelled from the characteristic polynomial?

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How can you show that t^4 – 3 (t^3) + 4 (t^2) – 3 t + 1 >= 0 for t > 0? - Quora
Answer (1 of 3): The easiest, and quickest, way is to graph the function and look for any points below 0. Use x instead of t, and set the function = y, not >=0. The function is a parabola with a minimum at (1,0). . If y goes below 0, the the expression is false. It is 0 at 1 but does not go below...