What is the value of A when we rewrite 4^31^x as A^x

Mathematics

1 answer:

Mumz [18]3 years ago

7 0

Answer:

This looks simple I belive the answer would be 35 because 4^x + 31^x = 35^x?

I might be wrong here.

Step-by-step explanation:

Hope this helps, have a good day! :)

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allochka39001 [22]

Answer:

आपको अपने पापों के लिए पश्चाताप करना चाहिए । वे तुम्हें नष्ट कर देंगे । आप कोई मौका नहीं खड़े हैं । तुम उसके हाथ पर मर जाओगे । पछताना

Step-by-step explanation:

its just so simple! :D

6 0

3 years ago

Read 2 more answers

AB and BC form a right angle at their point of intersection, B. If the coordinates of A and B are (14, -1) and (2, 1), respectiv

Studentka2010 [4]

Based on our conversation above, we can then easily find the missing x-coordinate. If the equation for line BC is y = 6*x - 11 and we know that the y-coordinate is 13, then

13 = 6*x - 11
24 = 6*x
4 = x

The x-coordinate is 4.

7 0

3 years ago

Help me with this question please

stich3 [128]

Answer:

(-1,9)

Step-by-step explanation:

union contains both sets.

4 0

3 years ago

A fraction becomes 4÷5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and den

tia_tia [17]

Answer:

$\dfrac{7}{9}$

Step-by-step explanation:

$\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1$

$\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5$

Putting it in matrix form

$\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}$

From Cramer's rule we have

$x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7$

$y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9$