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3 years ago

(e) the equations of the asymptotes (enter your answers as a comma-separated list of equations.)

Mathematics

1 answer:

nexus9112 [7]3 years ago

4 0

Answer:

x=3,x=1

y=2,y=-2

Step-by-step explanation:

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In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

3 0

3 years ago

Read 2 more answers

What property is used to show that 6(x + 7) = 6x + 42?

Ymorist [56]

Answer: Distributive because 6 has to be multiplied by x and 7

6 0

3 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .