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3 years ago

13

Which of the following best describes the equation below? A. linear B. both linear and nonlinear C. neither linear nor nonlinear

D. nonlinear

1 answer:

Julli [10]3 years ago

4 0

Answer:

Where is the equation at

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Describe the key aspects of the function f(x) = –x2 – 2x – 1.

amid [387]

Answer:

x-int of -1; y-int of -1; parabola that is reflected along the x-axis.

3 0

2 years ago

The graphs of f(x) and g(x) are shown below.

nexus9112 [7]

Answer:

On a coordinate plane, a straight line with a positive slope goes through points (negative 2, negative 6), (0, 0), and (2, 6).

Step-by-step explanation:

The function is given by f(x) = - x and another function is given by g(x) = 2x.

Now, (g - f)(x) will be given by

(g - f)(x) = 2x - (- x) = 3x .......... (1)

Now, we have to choose from the options of the graphs that is applicable to function (1).

So, on a coordinate plane, a straight line with a positive slope goes through points (negative 2, negative 6), (0, 0), and (2, 6). (Answer)

7 0

2 years ago

Read 2 more answers

X^2+20=2x what are the values of x?

morpeh [17]

Answer:

No real solutions.

Step.-by-step explanation:

Step 1: Subtract 2x from both sides.

For this equation: a=1, b=-2, c=20

Step 2: Use quadratic formula with a=1, b=-2, c=20.

3 0

2 years ago

Read 2 more answers

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. Th

denpristay [2]

Answer:

(1) Correct option is B.

(2) Correct option is C.

Step-by-step explanation:

The information provided is:

$n_{1}=200,\ \bar x_{1}=22.7,\ s_{1} = 4.5\\n_{2}=200,\ \bar x_{2}=19.7,\ s_{2} = 4.3$

The (1 - <em>α</em>)% confidence interval for the difference between two mean is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_[2}-2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} }$

The critical value of <em>t</em> is:

$\alpha /2=0.05/2=0.025$

degrees of freedom $=n_{1}+n_{2}-2=200+200-2=398$

$t_{\alpha/2, n_{1}+n_{2}-2}=t_{0.025, 398}=1.96$

Compute the 95% confidence interval for the difference between two mean as follows:

$CI=22.7-19.7\pm 1.96\sqrt{\frac{4.5^{2}}{200}+\frac{4.3^{2}}{200} }\\=3\pm0.8624\\=(2.1376, 3.8624)\\\approx(2.14, 3.86)$

Thus, the 95% confidence interval, (2.14, 3.86) implies that the true mean difference value is contained in this interval with probability 0.95.

Correct option is B.

The null value of the difference between means is 0.

As the value 0 is not in the interval this implies that there is a difference between the two means, concluding that priming does have an effect on scores.

Correct option is C.

4 0

2 years ago

A jet is coming in for a landing. It's currently 25,000 feet in

Mariana [72]

The angle of depression is 29.0521°. So it is a safe landing.

Step-by-step explanation:

Step 1:

The plane is flying at a height of 25,000 feet and 45,000 feet away from the landing strip. Assume it lands with an angle of depression of x°.

So a right-angled triangle can be formed using these measurements. The triangle's opposite side measures 25,000 feet while the adjacent side measures 45,000 feet. The angle of the triangle is x°.

To determine the value of x, we calculate the tan of the given triangle.

$tanx = \frac{opposite side}{adjacent side}.$

Step 2:

The length of the opposite side = 25,000 feet.

The length of the adjacent side = 45,000 feet.

$tan x = \frac{25,000}{45,000} = 0.5555.$

So x = 29.0521°. Since x < 30°, it is a safe landing.

8 0

3 years ago