All categories

3 years ago

Find the interval in which the function is negative.

Mathematics

1 answer:

DedPeter [7]3 years ago

3 0

Answer:

B. II only . . . (-1, 4)

Step-by-step explanation:

The function factors as ...

f(x) = (x +1)(x -4)

The factors are zero when x=-1 and x=4. The value of the function will be negative when one of these factors is negative, between x=-1 and x=4. In interval notation, that interval is ...

(-1, 4) . . . interval for f(x) < 0 . . . . . . interval description II (only)

You might be interested in

Question 5

sergey [27]

Explanation : If 66 is a factor of this unknown value, then 22 must be as well considering that 22 is a factor off 66. Let's say that this large value is 330. It is a multiple of 66, as 66 $*$ 5 = 330. At the same time 22 $*$ 15 = 330, so 330 is a multiple of 22 as well - or vice versa, 12 is a factor of 330.

We can also tell that 15, 22 fit into 330 through another approach. 22 $*$ 3 = 66, and 66 $*$ 5 = 330, so 5 $*$ 3 = 15 - the same value. This proves that 22 will always be a factor of a value that is the factor of 66.

3 0

3 years ago

Read 2 more answers

What are types of legislature

gulaghasi [49]

Answer:

Two common types of legislature are those in which the executive and the legislative branches are clearly separated, as in the U.S. Congress, and those in which members of the executive branch are chosen from the legislative membership, as in the British Parliament.

8 0

4 years ago

For what value of k are there two distinct real solutions to the original quadratic equation (k+1)x²+4kx+2=0.

lina2011 [118]

Answer:

k ∈ (-∞, $-\frac{1}{2}$ )∪(1,∞)

Step-by-step explanation:

For quadratic equations $ax^2+bx+c=0,a\neq 0$ you can find the solutions with the Bhaskara's Formula:

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\\and\\x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

A quadratic equation usually has two solutions.

If you only want real solutions the condition is that the discriminant ( $\Delta$ ) has to be greater than zero, this means:

$\Delta=b^2-4ac>0$

Then we have the expression:

$(k+1)x^2+4kx+2=0$

$a=(k+1)\\b=4k\\c=2\\$

Now to find two distinct real solutions to the original quadratic equation we have to calculate the discriminant:

$b^2-4ac>0\\(4k)^2-4.(k+1).2>0\\16k^2-8(k+1)>0\\16k^2-8k-8>0$

We got another quadratic function.

$16k^2-8k-8>0$ we can simplify the expression dividing both sides in 8.

$16k^2-8k-8>0\\\\\frac{16k^2}{8} -\frac{8k}{8} -\frac{8}{8} >\frac{0}{8}\\\\2k^2-k-1>0$

We can apply Bhaskara's Formula except that the condition in this case is that the solutions have to be greater than zero.

$2k^2-k-1>0\\a=2\\b=-1\\c=-1$

$k_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1+\sqrt{9} }{4}=\frac{1+3}{4} =1 \\and\\k_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.(-1)}}{2.2}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$

Then,

$k>1 \\and\\k$

The answer is:

For all the real values of k who belongs to the interval:

(-∞, $-\frac{1}{2}$ )∪(1,∞)

there are two distinct real solutions to the original quadratic equation $(k+1)x^2+4kx+2=0$

4 0

4 years ago

Find 2^30+2^29/2^31+2^30 number system

diamong [38]

Step-by-step explanation:

2³⁰+2²⁹÷2³¹+2³⁰

=2³⁰+2‐²+2³⁰

=2²(2²⁸+2‐⁴+2²⁸)

7 0

3 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

SIZIF [17.4K]

C = 4M + 5
4M = C - 5
M = (C - 5)/4

answer is D. Last one.

6 0

4 years ago