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3 years ago

Please help asap Will mark brainliest

Mathematics

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

(4,-4)

Step-by-step explanation:

(x,y)

y=3/4x - 7

y=3/4*(4)-7

y=3-7

y=-4

(4,-4)

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Please help me ASAP!!!!!!!!!!!!!!

earnstyle [38]

Answer:

7 Celsius

Step-by-step explanation:

Initial temperature at 12pm is 1 Celsius.

Rises by 6 (add by 6).

3 0

3 years ago

The quadratic function fhas a vertex at (3,4) and opens upward. The quadratic function g is shown below.

Alecsey [184]

Answer:

Option (C)

Step-by-step explanation:

Equation of the quadratic function having vertex at (3, 4) and opening upwards,

So the the minimum point of the function is (3, 4).

Therefore, minimum value of the function is 4 at x = 3.

y = (x - h)² + k [Here, (h, k) is the vertex]

g(x) = 2(x - 4)² + 3

Vertex of the parabola is (4, 3).

Since, leading coefficient is positive, parabola will open upwards.

Therefore, vertex will be the minimum point.

Minimum value of the function will be 3 at x = 4.

Minimum value of the function 'f' is greater than the minimum value of the function 'g'.

Option (C) will be the answer.

6 0

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

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

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