All categories

3 years ago

What is the undefined value of 3b^2+13b+4/b+4?

2 answers:

8 0

Let f(b) = <span>3b^2+13b+4/b+4
</span>the domain of f is D = {b/ b+4#0} so b# - 4

so the undefined value of is when b = -4

xxTIMURxx [149]3 years ago

6 0

The answer is D.

The rest of the answers are

1. A
2. B
3. A
4. D
5. B
6. D
7. A
8. B
9. C
10. D
11. C
12. C

You might be interested in

Can someone help me with 3

Artemon [7]

6 ft per minute is the answer

3 0

4 years ago

Read 2 more answers

Consider the system of quadratic equations \begin{align*} y &=3x^2 - 5x, \\ y &= 2x^2 - x - c, \end{align*}where $c$ is

shtirl [24]

Hello, we need to solve this system, c being a real number.

$\begin{cases}y &= 3x^2-5x\\y &= 2x^2-x-c\end{cases}$

y=y, right? So, it comes.

$3x^2-5x=2x^2-x-c\\\\3x^2-2x^2-5x+x+c=0\\\\\boxed{x^2-4x+c=0}$

We can compute the discriminant.

$\Delta=b^2-4ac=4^2-4c=4(4-c)$

If the discriminant is 0, there is 1 solution.

It means for $4(4-c)=0 4-c=0 \boxed{c=4}$

And the solution is

$x_2=x_1=\dfrac{4}{2}=2$

If the discriminant is > 0, there are 2 real solutions.

It means 4(4-c) > 0 <=> 4-c > 0 <=> $\boxed{c$

And the solution are

$x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{4-c}}{2}=2-\sqrt{4-c}\\\\x_2=2+\sqrt{4-c}$

If the discriminant is < 0, there are no real solutions.

It means 4(4-c) < 0 <=> 4-c < 0 <=> $\boxed{c>4}$

There are no real solutions and the complex solutions are

$x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{i^2(c-4)}}{2}=2-\sqrt{c-4}\cdot i\\\\x_2=2+\sqrt{c-4}\cdot i$

Thank you.

8 0

4 years ago

Need the answers before 8pm

saveliy_v [14]

Answer:

A reflection over the x- axis would be the correct answer.

Step-by-step explanation:

When you flip figure A over the x-axis it will land onto figure B.

5 0

4 years ago

TOY has coordinates T (-3, 4), O (-4, 1) and Y (-2, 3). a translation maps point T to T' (-1, 1). find the coordinates of O' and

NeX [460]

Answer:

Answer: The correct option is (A) O' (−2, −2); Y' (0, 0).

Step-by-step explanation: Given that the co-ordinates of the vertices of ΔTOY are T(−3, 4), O (−4, 1), and Y (−2, 3). A translation maps point T to T' (−1, 1).

We are to find the co-ordinates of the points O' and Y'.

The given transformation from T to T' is

T(−3, 4) ⇒ T' (−1, 1).

Let, (−3 + x, 4 + y) = (-1, 1).

So,

and

That is, the transformation rule is

(a, b) ⇒ (a+2, b-3).

Therefore,

co-ordinates of O' are (-4+2, 1-3) = (-2, -2),

and

co-ordinates of Y' are (-2+2, 3-3) = (0, 0).

Thus, the required co-ordinates of O' and Y' are (-2, -2) and (0, 0) respectively.

Option (A) is correct.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B50%5C%5C%5Csqrt%7B6%5C%5C%7D%20%7D" id="TexFormula1" title="\sqrt{50\\\sqrt{6\\} }"

satela [25.4K]

That the answer I’m pretty sure

5 0

3 years ago