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slamgirl [31]
3 years ago
7

I need helppppppppppppp please

Mathematics
1 answer:
netineya [11]3 years ago
5 0

Answer:

I want to say A)

Step-by-step explanation:

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HURRY PLS 15 MINS LEFT!!
Amiraneli [1.4K]

Answer:

y is less than or equal to 2/3x + 1/5

Step-by-step explanation:

7 0
2 years ago
At Southside Middle school, 30% if the students play an instrument. There are 310 students. How many students play an instrument
Svetach [21]
30% converted to a decimal is 0.30
310 x 0.3 = 93 students play an instrument
3 0
3 years ago
What is the answer to this
Pavel [41]

Answer: A

explanation: the vertex in this graph is the lowest point, making it a minimum vertex. If it was the highest point, it would be maximum

6 0
3 years ago
Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17
V125BC [204]

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}

Cramer's rule says that

x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}

where x_i is the solution for i-th variable, and \mathbf A_i is a modified version of \mathbf A with its i-th column replaced by \mathbf b.

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}

\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}

\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))

\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120

The modified matrices and their determinants are

\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240

\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360

\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480

Then by Cramer's rule, the solution to the system is

x = \dfrac{-240}{-120} \implies \boxed{x = 2}

y = \dfrac{360}{-120} \implies \boxed{y = -3}

z = \dfrac{-480}{-120} \implies \boxed{z = 4}

5 0
2 years ago
Number of learners in a class of 45 if x are absent ​
GalinKa [24]

Answer:

45 - x

Step-by-step explanation:

Number of learners is number in class minus those absent , that is

number of learners = 45 - x

7 0
3 years ago
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