How many solutions does the nonlinear system of equations graphed below have?

Mathematics

1 answer:

defon2 years ago

8 0

Answer:

There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.

Step-by-step explanation:

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Answer:

$6A- 8B = \left[\begin{array}{ccc}56&50&-14\\-126&92&62\\-78&26&2\end{array}\right]$

Option A is correct.

Step-by-step explanation:

$A=\left[\begin{array}{ccc}-4&-1&-9\\-9&6&9\\-1&-1&3\end{array}\right] \\B=\left[\begin{array}{ccc}-10&-7&-5\\9&-7&-1\\9&-4&2\end{array}\right]$

We need to find 6A - 8B

Multiply matrix A with 6 and matrix B with 8 and then subtract 6A - 8B

$6A=\left[\begin{array}{ccc}-24&-6&-54\\-54&36&54\\-6&-6&18\end{array}\right] \\8B=\left[\begin{array}{ccc}-80&-56&-40\\72&-56&-8\\72&-32&16\end{array}\right]\\6A - 8B =\left[\begin{array}{ccc}-24-(-80)&-6-(-56)&-54-(-40)\\-54-(72)&36-(-56)&54-(-8)\\-6-(72)&-6-(-32)&18-(16)\end{array}\right] \\6A - 8B =\left[\begin{array}{ccc}-24+80&-6+56&-54+40\\-54-72&36+56&54+8\\-6-72&-6+32&18-16\end{array}\right] \\6A- 8B = \left[\begin{array}{ccc}56&50&-14\\-126&92&62\\-78&26&2\end{array}\right]$