All categories

3 years ago

Which value is in the domain of f(x)?

Mathematics

2 answers:

Vlad1618 [11]3 years ago

6 0

Answer:

Step-by-step explanation:

The domain is the inputs (or the x values)

We start at -6 (but do not include it) and end at +4 (we include it)

-6 < x ≤4

The value that is included is 4

ahrayia [7]3 years ago

4 0

Answer:

Step-by-step explanation:

$f(x)=-2x+3, 0$

You might be interested in

Given that the sec A = 13/5 and A is in QIV, determine exact value for the sin A.

AysviL [449]

Answer:

The exact value of sin A is opp/hyp = -12/13

Step-by-step explanation:

If A is in Quadrant IV then the "opposite side" is negative and the "adjacent side" is positive. Since sec A = hyp / adj = 13 / 5 and since opp^2 + adj^2 must add up to hyp^2, Therefore y = √(hyp^2 - adj^2), or

y = -√(13^2 - 25) = -√144 = 12.

The exact value of sin A is opp/hyp = -12/13

6 0

3 years ago

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .

Verify that $(x + 2)\, (x + 3)$ indeed expands to the original polynomial:

$\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}$ .

4 0

3 years ago

The table shows the temperature of an amount of water set on a stove to boil, recorded every half minute.

creativ13 [48]

According to the given table the line of best fit is

f(x)=4.54x+77.84

The term "line of best fit" describes a line that passes across a scatter plot of data points and best captures their connection. Statisticians often utilise regression analysis software or manual computations to arrive at the geometric equation for the line using the least squares approach. A straightforward linear regression study of two or more independent variables will provide a straight line.

Hence, as asked by the question we need to find the time at which the tempurature will become 100°C.

To find that let us put f(x)=100 and find out the value of x for which it is satisfied.

4.54x+77.84=100

⇒ $x=\frac{100-77.84}{4.54}=4.88$

⇒x≈5

Therefore the time at which the tempurature is 100°C is 5 minutes.

Learn more about "line of best fit" here-

brainly.com/question/14279419

#SPJ10

4 0

2 years ago

Answer for me, I'm on a big test

sergejj [24]

Answer:

D is your answer because you need to multiply power of a with 4. Above there is image you can understand.

4 0

3 years ago

Need help with 5 and 6. Please explain steps. Thanks!!!!

alukav5142 [94]

5. Answer: rate of 6 mph + current of 5 mph = 11 mph

<u>Step-by-step explanation:</u>

Use d = r * t

$\begin {array}{l|c|c|c||l} &\underline{time}&\underline{rate}&\underline{distance}&\underline{equation}\\ with&x&r+5&33&x(r+5)=33\\against&6-x&r-5&3&(6-x)(r-5)=3\\\end{array} \\\\\\\text{Solve each equation for x.}\\\\x(r + 5) = 33\qquad \qquad \qquad (6-x)(r-5)=3\\\\.\qquad \quad x=\dfrac{33}{r+5}\qquad \qquad \qquad \qquad 6-x=\dfrac{3}{r-5}\\\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x=-\dfrac{3}{r-5}+6$

$\text{Now, set the equations equal to each other and solve for r:}\\\\\dfrac{33}{r+5}=-\dfrac{3}{r-5}+6\\\\\\\dfrac{33}{r+5}=-\dfrac{3}{r-5}+\dfrac{6(r-5)}{(r-5)}\\\\\\\dfrac{33}{r+5}=\dfrac{6r-33}{r-5}\\\\\\33(r-5)=(r+5)(6r-33)\\\\\\33r - 165 = 6r^2-3r-165\\\\\\.\qquad \quad 0=6r^2-36r\\\\\\.\qquad \quad 0=6r(r-6)\\\\\\.\qquad \quad 0=6r\quad or\quad 0=r-6\\\\\\.\qquad \quad r=0\quad or\quad r=6\\\\\text{0 mph is an erroneous solution so Karl's rate was 6 mph}$

6. Answer: 5 hours

<u>Step-by-step explanation:</u>

Use d = r * t. Since they both traveled the same distance, we can set the (rate * time) for each person equal to each other:

Wilbur: rate = 33 and time = x --> equation: 33x

Mary: rate = 55 and time = x - 2 --> equation: 55(x - 2)

33x = 55(x - 2)

33x = 55x - 110

0 = 22x - 110

110 = 22x

5 = x

8 0

4 years ago