Ask question

Ask question

All categories

2 years ago

15

Part 1: What mistake did AJ make in the graph?

1 answer:

grigory [225]2 years ago

8 0

Answer:

Part 1) AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) see the explanation

Step-by-step explanation:

Part 1) What mistake did AJ make in the graph?

we have

$f(x)=-(x+2)^2-1$

This is the equation of a vertical parabola written in vertex form

The parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The vertex is the point (-2,-1)

therefore

AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) Evaluate any two x-values (between -5 and 5) into AJ's function. Show your work. How does your work prove that AJ made a mistake in the graph?

take the values x=-4 and x=4

For x=-4

substitute the value of x in the quadratic equation

$f(x)=-(-4+2)^2-1\\f(x)=-5$

For x=4

substitute the value of x in the quadratic equation

$f(x)=-(4+2)^2-1\\f(x)=-37$

According to AJ's graph for the value of x=-4 the function should be positive, however it is negative and for the value of x=4 the function should be positive and the function is negative

therefore

AJ made a mistake in the graph

You might be interested in

How many 1/4 cup servings of raisins are in 7/8 cups of raisins?

Ostrovityanka [42]

Answer:

7

Step-by-step explanation:

Divide 7/8 by 1/4. (You divide the "supply" of raisins into 1/4 cup "parts".)

$\frac{7}{8} \div \frac{1}{4} = \frac{7}{8} \times \frac{4}{1} = \frac{28}{8} =\frac{7}{1} =7$

Multiply 7/8 by the reciprocal of 1/4 ("flip" the 1/4).

The result is 28/8, which can be simplified ("reduced") to 7.

3 0

2 years ago

Find the general solution of (x+3)y’=2y

gregori [183]

Answer:

$y=C(x+3)^2$

Step-by-step explanation:

We are given:

$\displaystyle (x+3)y^\prime=2y$

Separation of Variables:

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}$

So:

$\displaystyle \frac{dy}{y}=\frac{2}{x+3} \, dx$

Integrate:

$\displaystyle \int\frac{dy}{y}=\int\frac{2}{x+3}\, dx$

Integrate:

$\displaystyle \ln|y|=2\ln|x+3|+C$

Raise both sides to e:

$|y|=e^{2\ln|x+3|+C}$

Simplify:

$|y|=(e^{\ln|x+3|})^2\cdot e^C$

So:

$|y|=C|x+3|^2$

Simplify:

$y=\pm C(x+3)^2=C(x+3)^2$

5 0

3 years ago

Factor completely 3x^2 - x - 4

kakasveta [241]

Answer:

<em>B. </em> $(3x-4)(x+1)$ <em></em>

Step-by-step explanation:

6 0

3 years ago

Find the percent of the area under the density curve where x is more than 3. <br>

Gekata [30.6K]

Answer:

The percent of the area under the density curve where $x$ is more that 3 is 25 %.

Step-by-step explanation:

Since the density curve is a linear function, the area under the curve can be calculated by the geometric formula for a triangle, defined by the following expression:

$A = \frac{1}{2}\cdot (x_{f} - x_{o})\cdot (y_{f}-y_{o})$ (1)

Where:

$A$ - Area, in square units.

$x_{f} - x_{o}$ - Base of the triangle, in units.

$y_{f} - y_{o}$ - Height of the triangle, in units.

The percent of the area is the ratio of triangle areas under the density curve multiplied by 100 per cent, that is:

$x = \frac{\frac{1}{2}\cdot (5-3)\cdot (0.25) }{\frac{1}{2}\cdot (5-1)\cdot (0.5) }\times 100\,\%$

$x = 25\,\%$

The percent of the area under the density curve where $x$ is more that 3 is 25 %.

7 0

3 years ago

A package contains 12 resistors, 3 of which are defective. If 4 are selected, find the probability of getting

s344n2d4d5 [400]

Answer:

Incomplete question, but I gave a primer on the hypergeometric distribution, which is used to solve this question, so just the formula has to be applied to find the desired probabilities.

Step-by-step explanation:

The resistors are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

12 resistors, which means that $N = 12$

3 defective, which means that $k = 3$

4 are selected, which means that $n = 4$

To find an specific probability, that is, of x defectives:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = x) = h(x,12,4,3) = \frac{C_{3,x}*C_{9,4-x}}{C_{12,4}}$

7 0

3 years ago