What is the slope of y-9=-2(x-8)

Answer:

[tex]y\leq \frac{1}{3}x-4[/tex]

Step-by-step explanation:

we know that

The solution of the inequality is the shaded area below the solid line

The slope of the line is positive

The y-intercept of the solid line is equal to [tex]-4[/tex]

therefore

The inequality must be

[tex]y\leq \frac{1}{3}x-4[/tex]

Answer:

The correct answer is f(n)=3n+38

Step-by-step explanation:

Put in y=mx+b form, so y= f(n), $3 is fee per video game rented and n is the number of video games rented, b is ($18 fixed fee for this year +$20 fee for next year). So, f(n)=3n+38.

Answer:

(C)7x-3x

Step-by-step explanation:

The given algebraic expression is:

"The difference between seven times a number and three times that number"

Let the number be equal to =x, then according to the statement, we have

seven times a number=7x and three times that number=3x

The difference between seven times a number and three times that number=7x-3x

which is the required algebraic expression.

Hence, option C is correct.

Answer:

Step-by-step explanation:

Answer:  The required co-ordinates of the point S are (-11, -6).

Step-by-step explanation:  We are given to find the co-ordinates of the point S that lies along the directed line segment from R(-14, -1) to T(4, -13) and partitions the segment in the ratio of 1:5.

We know that

the co-ordinates of a point that divides the line segment joining the points (a, b) and (c, d) in the ratio m : n is given by

[tex]\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).[/tex]

According to the given information, we have

(a, b) = (-14, -1),  (c, d) = (4, -13)  and  m : n = 1 : 5.

Therefore, the co-ordinates of the point S are given by

[tex]\left(\dfrac{1\times4+5\times(-14)}{1+5},\dfrac{1\times(-13)+5\times(-1)}{}\right)\\\\\\=\left(\dfrac{4-70}{6},\dfrac{-13-5}{6}\right)\\\\\\=\left(\dfrac{-66}{6},\dfrac{-18}{6}\right)\\\\=(-11,-3)[/tex]

Thus, the required co-ordinates of the point S are (-11, -6).

Answer: three hundred fourteen thousand, two hundred seven

explanation for the comma after thousand: I’m used to putting one there because it says 314, not just 314 so I added one. Hope I helped !

Hello : let  A(2,-1)    B(8,4)
the slope is :   (YB - YA)/(XB -XA)
(4+1)/(8-2)  = 5/6

Answer: Equation of line in point slope form,

[tex]y + 1 = 5 ( x - 2 )[/tex]

And, Equation of line in standard form,

[tex]5 x - 6 y = 16[/tex]

Step-by-step explanation:

Since, If a line passes through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] ,

Then the equation of line,

[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1} (x-x_1)[/tex]

Here [tex]x_1 = 2[/tex], [tex]y_1=-1[/tex], [tex]x_2=8[/tex] and [tex]y_2=4[/tex]

Thus, the equation of the given line,

[tex]y-(-1)=\frac{4-(-1)}{8-2} (x-2)[/tex]

⇒ [tex]y+1=\frac{4+1}{8-2} (x-2)[/tex]

⇒ [tex]y+1=\frac{5}{6} (x-2)[/tex] -----(1)

⇒  [tex]6(y+1)= 5(x-2)[/tex]

⇒ 6 y + 6 = 5 x - 10

⇒ 6 = 5x - 6y - 10 ( By subtracting by on both sides )

⇒ 6 + 10 = 5x - 6y  ( By adding 10 on both sides )

⇒ 16 = 5x - 6y

⇒ 5 x - 6 y = 16 ------(2)

Since, in slope for of a line is, [tex]y-y_1= m (x-x_1)[/tex]

Thus, equation (1) shows the in slope form of the line.

And, standard form of the line is ax + by = c where a, b and c are the integers.

Thus, equation (2) shows the standard form of the given line.

Answer:

Sequence 1 is arithmetic and Sequence 2 is geometric.

The feasible vertices of the system representing Josh's exam are (0,0), (12,5), and (15,1.25). The non-feasible vertices are (15,0), (0,5), and (20,0). Therefore, the ordered pair (15,0) is not a vertex of the feasible region.

1. System of Inequalities:

  - Time constraint: [tex]\(2m + 8s \leq 40\)[/tex]

  - Number of multiple-choice questions: [tex]\(m \leq 15\)[/tex]

  - Number of short-answer questions: [tex]\(s \leq 5\)[/tex]

2. Feasible Vertices:

  - (0,0), (12,5), (15,1.25)

3. Non-feasible Vertices:

  - (15,0), (0,5), (20,0)

So, the ordered pair (15,0) is not a vertex of the feasible region.

Squaring both sides of an inequality can be valid if you know both sides are non-negative, but it can introduce extraneous solutions. Therefore, it's important to check any solutions against the original inequality to ensure their validity, especially in cases involving negative numbers or functions with restricted domains.

When you're working with inequalities, you have to be careful when performing operations like squaring both sides. Unlike equalities, where multiplication or division by the same number on both sides does not change equality, with inequalities, the effect can be more complex due to the direction of the inequality sign and the possibility of dealing with negative numbers.

For instance, squaring both sides of an inequality is not always a valid operation because if one or both sides of the inequality are negative, squaring could lead to incorrect results. When you square a negative number, it becomes positive, which could potentially reverse the inequality's direction. However, if you know that both sides of the inequality are non-negative, then squaring both sides is permissible. This concept is similar to solving quadratic constraints without introducing square roots, using identities like |(1 + ix)²|² = ([1 + ix|²)².

To avoid introducing solutions that were not there originally (extraneous solutions), it is important to check the solutions obtained after squaring against the original inequality. An example where squaring both sides might be questioned is when solving trigonometric inequalities, where a common mistake is to square both sides without considering the domain of the original function.

Answer: I think the answer is A.

Step-by-step explanation:

I think this because its the only time that It could stay the same.

From the information, A is a right angle, B is an acute triangle and C is an acute angle.

It will be a right triangle if a² + b² = c². It will be аcute if a² + b² > c² and it'll be obtuse if a² + b² < c².

For the first one,

a² + b² = 3² + 4² = 9 + 16 = 25 and c² = 5² = 25

25 = 25

This is a right triangle.

For the second one,

a² + b² = 5² + 6²

= 25 + 36 = 61

c² = 7² = 49

61 > 49 = аcute triangle.

For the third one,

a² + b² = 8² + 9²

= 64 + 81 = 145

c² = 12² = 144

145 > 144 = аcute triangle.

Learn more about triangles on:

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Answer:

Option 3th is correct

x = 0.5

Step-by-step explanation:

The given functions are:

Function 1: [tex]y = 4^x[/tex]

Function 2: [tex]y=7x^2+4x-2[/tex]

The values of function at x = -0.5, 0, 0.5 and 2 are as follow:

x values     Function 1              Function 2

-0.53                 0.5                            -2.25

0                        1                              -2

0.53                   2.086                      2.0863

2                        16                             34    

From the above table. It is clear that the quadratic function [tex]y=7x^2+4x-2[/tex]  exceeds the exponential function [tex]y = 4^x[/tex] at x = 0.53

Therefore, the approximate x-value in which the quadratic function exceeds the exponential function at x = 0.5

Answer: a) y = 90-15x

b) y = 90

c) x = 6

Step-by-step explanation:

Since we have given that

Amount borrowed from his grandmother = $90

Amount he pay back each week = $15

Let the number of weeks be 'x'.

Let total amount owed be 'y'.

So, it becomes

[tex]y=90-15x[/tex]

a) write an equation in standard for that represents the amount owed y after x weeks.

Our equation is [tex]y=90-15x[/tex]

b. find the y intercept of the graph. what does this represent?

For this , we put x = 0.

[tex]y=90-15(0)\\\\y=90[/tex]

So, the y-intercept of the graph is 90.

It represents that the graph intersect y-axis at (0,90)

c. find the x intercept of the graph. what does this represent?

For this, we put y = 0.

[tex]0=90-15x\\\\90=15x\\\\x=\dfrac{90}{15}\\\\x=6[/tex]

So, the x intercept of the graph is 6.

It represents that the graph intersect the x-axis at (6,0)

The left-most point on the number line is 10.

Minimum value = 10

The right-most point is 45.

Maximum value= 45

First quartile or lower quartile = 16

Third quartile or upper quartile = 36

The median is 28.

The box plot shows the total amount of time, in minutes, the students in a class spend traveling to school each day:

A box plot is titled Daily Travel Time and labeled Time (min).

The left-most point on the number line is 10 and the right-most point is 45.

The box is labeled 16 on the left edge and 36 on the right edge.

A vertical line is drawn inside the rectangle at point 28.

The whiskers are labeled as 11 and 43.

What information is provided by the box plot?

In the box plot, the information is given as follows.

         Minimum value = 10

        Maximum value= 45

To know more about Median click the given below.

https://brainly.com/question/9969858

The 1st quartile q1 is also called the 25th percentile (the bottom 25% of a data set). In this case, if we choose 100 light bulbs, then the 1st quartile will be the 25 bulbs with shortest useful life.

To solve this problem, we would have to use the z statistic. Using the standard distribution tables for z, we locate for the value of z at P = 0.25:

z = - 0.674

Then to calculate for the 1st quartile, we use the formula:

Q1 = x + z * s

where x is the mean, and s is the standard deviation, therefore:

Q1 = 392 + (- 0.674) (9)

Q1 = 385.934

To find the first quartile of a normal distribution, use the z-score corresponding to an area of 25%. For these lightbulbs with a mean of 392 hours and standard deviation of 9 hours, the first quartile is approximately 386 hours.

In statistics, the quartiles of a dataset divide the data into four equal parts. The first quartile, Q1, is the value below which 25% of the data fall. To determine Q1 for the lifetime of a certain type of lightbulb, we use the properties of the normal distribution.

First, realize that 25% of data falls below Q1, and hence using the z-table, we find that z=-0.675 corresponds to an area of 0.25. We then use the formula for a z-score in a normal distribution, z = (X - μ)/σ. Solving for X we find X = z*σ + μ.

Plugging in the known values we get Q1 = -0.675*9 + 392 = 386.025 hours. So, for these lightbulbs, 25% will have failed by around 386 hours.

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