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2 years ago

Help, please I'm taking a test on plato

Mathematics

2 answers:

ratelena [41]2 years ago

5 0

Answer:mark me as brainliest plz

Step-by-step explanation:

it's c

STALIN [3.7K]2 years ago

5 0

Answer:

Your answer is C, x = 3 and y = -6.

Step-by-step explanation:

Hope this helps!

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If 32^2c= 8^c+7, what is the value of c

Leona [35]

$\bf 32^{2c}=8^{c+7}~~ \begin{cases} 32=2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^5\\ 8=2\cdot 2\cdot 2\\ \qquad 2^3 \end{cases}\implies (2^5)^{2c}=(2^3)^{c+7} \\\\\\ 2^{5(2c)}=2^{3(c+7)}\implies \stackrel{\textit{if the bases are the same, then so are the exponents}}{5(2c)=3(c+7)\implies 10c=3c+21} \\\\\\ 7c=21\implies c=\cfrac{21}{7}\implies c=3$

3 0

2 years ago

Read 2 more answers

This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago

Give me the answers to thissss plssss rnnnn asapppp!!!!

soldier1979 [14.2K]

$\\ 1. (3x ^{2} + 2xy + 7) - (6x ^{2} - 4xy + 3) \\ = 3x ^{2} + 2xy + 7 - 6x ^{2} + 4xy - 3 \\ = 3x ^{2} - {6x}^{2} + 2xy + 4xy + 7 - 3 \\ = { - 3x}^{2} + 6xy + 4 \\ 2. \: {9x}^{2} - 2x + 3 - ( {4x}^{2} + 7x - 5) \\ = 9x ^{2} - 2x + 3 - 4x ^{2} - 7x + 5 \\ = {9x}^{2} - {4x}^{2} - 2x - 7x + 3 + 5 \\ = {5x}^{2} - 9x + 8 \\ 3. \: {x}^{2} - 5x + 6 \\ = {x}^{2} - (3 + 2)x + 6 \\ = {x}^{2} - 3x - 2x + 6 \\ = x(x - 3) - 2(x - 3) \\ = (x - 3)(x - 2) \\ hope \: it \: helps \\ good \: luck \: on \: your \: assignment$