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2 years ago

Solve the inequality and enter your solution as an inequality in the box below, using "" for 2 if necessary. -5(4x + 7)

Mathematics

1 answer:

kvasek [131]2 years ago

6 0

Answer:

I simplified it to -20x - 35

Step-by-step explanation:

-5 times 4x is -20x, -5 times 7 is -35.

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Step-by-step explanation:

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For the polynomial function ƒ(x) = x4 − 12x3 + 46x2 − 60x + 25, find all local and global extrema.

rodikova [14]

$\bf f(x)=x^4-12x^3+46x^2-60x+25\\\\
-------------------------------\\\\
\cfrac{df}{dx}=4x^3-36x^2+92x-60\implies 0=4x^3-36x^2+92x-60
\\\\\\
0=x^3-9x^2+23x-15\impliedby \textit{doing some synthetic division with 5}
\\\\\\
0=(x^2-4x+3)(x-5)\implies 0=(x-3)(x-1)(x-5)\\\\
-------------------------------\\\\
\begin{cases}
f(1)=0\leftarrow minimum&1,0\\
f(3)=16\leftarrow maximum&3,16\\
f(5)=0\leftarrow minimum&5,0
\end{cases}\impliedby \textit{3 extrema, locals and globals}$

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2 years ago

Money Flow An investment scheme is expected to produce a continuous flow of money, starting at $1000 and increasing exponentiall

kvasek [131]

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Step-by-step explanation:

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2 years ago

Express the system as AX = B; then solve using matrix inverses

Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

$\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B$

Adj.A=Transpose of cofactor of Matrix A

$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Solve the inequality and enter your solution as an inequality in the box below, using "" for 2 if necessary. -5(4x + 7) ​

Solve the inequality and enter your solution as an inequality in the box below, using "" for 2 if necessary. -5(4x + 7)