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Sati [7]
3 years ago
15

Please help

Mathematics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

the

Step-by-step explanation:

sale would be $20 for every item, duh

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Help me to do this problem System by sustitution. <br>4) y = -3x - 7 <br>-6x - 2y = 14​
valentina_108 [34]

Answer:

The system of equations has infinite solutions

Step-by-step explanation:

∵ y = -3x - 7 ⇒ (1)

∵ -6x - 2y = 14 ⇒ (2)

→ Substitute y in equation (2) by equation (1)

∴ -6x - 2(-3x - 7) = 14

→ Simplify the left side

∵ -6x - 2(-3x) - 2(-7) = 14

∴ -6x + 6x + 14 = 14

→ Add the like terms in the left side

∵ (-6x + 6x) + 14 = 14

∴ 0 + 14 = 14

∴ 14 = 14

∵ The variable is disappeared and the two sides of the equation equal

∴ The equation has many solutions

→ That means x and y can be any numbers

∴ The system of equations has infinite solutions

7 0
3 years ago
Please anyone I’m desperate
Savatey [412]

Answer:

the answer is B) 12a+6b

Step-by-step explanation:

3+9=12 so 3a+9a=12a,

7-1=6 so 7b-(1)b=6b

3 0
3 years ago
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URGENT <br> WILL GIVE BRAINLIEST <br> solve using matrices<br> 4x + 5y = 40 <br> x – y = 1
dalvyx [7]

Answer:

x = 5, y = 4

Step-by-step explanation:

\left\{\begin{array}{ccc}4x+5y=40\\x-y=1\end{array}\right\\\\A=\left[\begin{array}{ccc}4&5\\1&-1\end{array}\right]\\\\detA=\left|\begin{array}{ccc}4&5\\1&-1\end{array}\right|=(4)(-1)-(1)(5)=-4-5=-9\\\\A^D=\left[\begin{array}{ccc}-1&-1\\-5&4\end{array}\right]\\\\\left(A^D\right)^T=\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]

A^{-1}=\dfrac{1}{detA}\left(A^D\right)^T\\\\A^{-1}=\dfrac{1}{-9}\left[\begin{array}{ccc}-1&-5\\-1&4\end{array}\right]=\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]

A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\A^{-1}A\cdot\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =A^{-1}\cdot\left[\begin{array}{ccc}40\\1\end{array}\right]

\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\dfrac{1}{9}&\dfrac{5}{9}\\\dfrac{1}{9}&-\dfrac{4}{9}\end{array}\right]\cdot\left[\begin{array}{ccc}40\\1\end{array}\right] \\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}\left(\frac{1}{9}\right)(40)+\left(\frac{5}{9}\right)(1)\\\\\left(\frac{1}{9}\right)(40)+\left(-\frac{4}{9}\right)(1)\end{array}\right]\\\\\left[\begin{array}{ccc}x\\y\end{array}\right] =\left[\begin{array}{ccc}5\\4\end{array}\right]\Rightarrow x=5,\ y=4

3 0
3 years ago
Please help with this question (the attached image)
Viktor [21]

Answer:

it is c

i had this test pls marek me brainlyesest

4 0
3 years ago
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True or False? When doing linear regression, if the correlation coefficient is positive, the slope of the line is negative.
skelet666 [1.2K]
The correct answer is false
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