How to find turning point of (c-2)^3+8

Answer:

b

Step-by-step explanation:

The greatest common factor is the largest value two numbers share in common which multiplies to each of the numbers.

7ab has factors 7, a, b

[tex]8b^3[/tex] has factors 2, 4, 8, b, b, b

These share in common b.

The solution for the system of equations is incorrect because line [1] should have a y-intercept at (0, 2) and not at (0, 1).

The general equation of a straight line is -

[y] = [m]x + [c]

where -

[m] is slope of line which tells the unit rate of change of [y] with respect to [x].

[c] is the y - intercept i.e. the point where the graph cuts the [y] axis.

The equation of a straight line can be also written as -

Ax + By + C = 0

By = - Ax - C

y = (- A/B)x - (C/A)

Given is the system of equations of straight lines -

y = -1/3x + 2

y = 2x - 5

plotted on the graph.

The error Emma made is that while plotting Equation [1] -

y = -1/3x + 2

The line should have passed from point (0, 2) on y - axis and not from (0, 1)

Therefore, the solution for the system of equations is incorrect because

Line [1] should have a y-intercept at (0, 2) and not at (0, 1).

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[Refer to the image attached for full question]

Final answer:

The weight shown on the deli scale for 5/10 pound of pepperoni is 0.5 pounds.

Explanation:

When Marco puts 5/10 pound of pepperoni on the deli scale, the scale which measures in hundredths of a pound will show this weight as 0.50 pounds.

This is because 5/10 can be simplified to 1/2, and when expressed as a decimal, 1/2 is equivalent to 0.5.

Therefore, the deli scale that reads to the hundredths place will display the weight as 0.50 pounds.

Answer:

Step-by-step explanation:

Using the formula for Grade Resistance/Force in this case...

!!AND MAKING SURE YOUR CALCULATOR IS IN DEGREE MODE!!

--> R/F: the Grade Resistance

--> W: Weight of the Vehicle

--> ∅: Angle of the Hill

Because the Grade Resistance in problem 1) is (-)130lbs... we know that the car is traveling DOWNhill. (If it were simply 130lbs, the car would be traveling up the hill)

to do this, we must take the INVERSE of sin.... (because we cannot simply divide both sides by 'sin' by itself... we must instead...

---> sin[tex]^{-1}[/tex](-0.05) = ?

∅ ≅ -2.866 ..... (you can then divide 866 by 60' if necessary, to find degree and minute answer)...

∅ ≅ -2° 14'

Hope this will help you and others with the 2nd question as well! ^_^

DONT GIVE UP! You got this '_^

The angle of the grade to the nearest tenth of a degree will be ∅ ≅ -2° 14'.

The branch of mathematics and physics known as mechanics is concerned with the interactions between matter, force, and motion in physical objects.

Using the formula for Grade Resistance/Force in this case...

R(or F) = Wsin∅

R/F: the Grade Resistance

W: Weight of the Vehicle

∅: Angle of the Hill

Because the grade resistance in problem 1) is (-)130lbs... we know that the car is traveling downhill. (If it were simply 130lbs, the car would be traveling up the hill)

W: 2600lbs       R: -130lbs      ∅: ??

(plugin) -->   -130lbs = 2600lbs(sin∅)

÷ both sides by 2600lbs

-130lbs÷2600lbs (lbs cancel) = -0.05

-0.05 = sin∅

When you get to this point. We have just found SIN∅.... but we were asked to find it ∅. To do this, we must take the inverse of sin.... (because we cannot simply divide both sides by 'sin' by itself... we must instead...

sin(-0.05) = ?

∅ ≅ -2.866 ..... (you can then divide 866 by 60' if necessary, to find the degree and minute answer)...

∅ ≅ -2° 14'

Therefore, the angle of the grade to the nearest tenth of a degree will be ∅ ≅ -2° 14'.

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Area of a triangle = [tex]\frac{1}{2} base*height[/tex]

In a right angle triangle the two legs are the base and height of the triangle.

One angle of the triangle is 23 degree  and one adjacent leg is 27.6 cm.Let the other leg opposite to 23 degree angle by x.Sine of an angle is ratio of opposite and hypotenuse. Hypotenuse is given as 30 cm.

[tex]Sin 23=\frac{x}{30}[/tex]

x= 30 sin23.

x=11.72cm.

The legs of the triangle are 11.72 cm and 27.6 cm.

Area of triangle =[tex]\frac{1}{2} x11.72 x27.6 =161.7cm^{2}.[/tex]

Final answer:

To find out how much more was cut from the skirt than the pants, convert fractions to have a common denominator and subtract: 3/4 inch minus 1/6 inch equals 7/12 inch.

Explanation:

To determine how much more the tailor cut off the skirt than the pants, we need to subtract the length cut from the pants from the length cut from the skirt. The tailor cut 3/4 of an inch off the skirt and 1/6 of an inch off the pants. To perform the subtraction, we need a common denominator, in this case, 12 is suitable. So we convert 3/4 to 9/12 and 1/6 to 2/12.

Now, we subtract the smaller fraction from the larger one:

9/12 - 2/12 = 7/12 of an inch.

Therefore, the tailor cut 7/12 of an inch more off the skirt than the pants.

The row of 9 bolts laid end to end would be 7 7/8 inches long.

To find the total length of 9 bolts laid end to end, we multiply the length of one bolt by the number of bolts.

Given:

- Length of one bolt: [tex]\( \frac{7}{8} \)[/tex] inch

- Number of bolts: 9

Total length:

[tex]\[ \text{Total length} = \text{Length of one bolt} \times \text{Number of bolts} \][/tex]

[tex]\[ \text{Total length} = \frac{7}{8} \times 9 \][/tex]

Now, multiply:

[tex]\[ \text{Total length} = \frac{7 \times 9}{8} \][/tex]

[tex]\[ \text{Total length} = \frac{63}{8} \][/tex]

To express this in mixed number form:

[tex]\[ \frac{63}{8} = 7 \frac{7}{8} \][/tex]

Therefore, the row of 9 bolts laid end to end would be [tex]\( 7 \frac{7}{8} \)[/tex] inches long.

The distribution of speeding tickets received by students is predicted to be right-skewed, with most students having no tickets and a few having multiple, indicative of a mode less than the median, which is less than the mean.

The question concerns the prediction of the shape of the distribution of speeding tickets received by 90 students who drive over the last year. The correct answer is c. the distribution will be​ right-skewed. Most individuals tend to obey traffic laws to avoid fines and penalties, meaning the majority of people will have no tickets.

However, there will always be a few individuals who, due to various reasons, might accumulate one or more tickets within a year. This creates a long tail to the right in the distribution, as we have a large number of students with 0 or very few tickets and a smaller number with higher counts of tickets. This results in the mode being less than the median, which is less than the mean, a characteristic of a right-skewed distribution.

The equation to represent the relationship between the time Sabar walks, denoted by [tex]x[/tex] , and the number of calories he burns, denoted by [tex]y[/tex], can be written as [tex]\( y = mx \),[/tex] where [tex]\( m \)[/tex] is the rate at which calories are burned per unit time.

To establish a relationship between the time Sabar spends walking and the number of calories he burns, we can use a simple linear equation. In this context, the independent variable [tex]x[/tex] represents the time spent walking, and the dependent variable [tex]\( y \)[/tex] represents the number of calories burned.

The rate at which calories are burned is typically constant for a given activity if the intensity remains consistent. This rate is represented by the coefficient [tex]\( m \)[/tex] in the equation [tex]\( y = mx \)[/tex]. The value of [tex]\( m \)[/tex] would be determined by how many calories Sabar burns per minute, hour, or any other time unit chosen for [tex]x.[/tex]

For example, if Sabar burns 5 calories per minute, then [tex]\( m = 5 \)[/tex]calories/minute, and the equation would be [tex]\( y = 5x \)[/tex]. If we measure time in hours and Sabar burns 300 calories per hour, then [tex]\( m = 300 \)[/tex]calories/hour, and the equation would be [tex]\( y = 300x \).[/tex]

This linear equation assumes that the relationship between time and calories burned is directly proportional, meaning that the number of calories burned increases at a constant rate as the time spent walking increases. This is a common assumption for steady-state aerobic activities like walking at a consistent pace.

In this exercise we will use the knowledge of statistics to calculate the value of the constant, so we have to:

[tex]k = 37.98 m^{-1}s^{-1}[/tex]

Recalling how the form of the rate equation is written, we have:

[tex]rate = k [a]^x[b]^n[/tex]

where:

Putting the already known values ​​into the given equation:

[tex]2.83 = k[0.273 m]*[0.763 m]^n \\2.83 = k [0.273 m]*[1.526 m]^n \\1 = [1.526/0.763]^n\\1 = 2^n\\n = 0[/tex]

Now trying to find the value of X in the given equation:

[tex]25.47 =k [0.819]*[0.763]^n\\2.83 = k[0.273]*[0.763]^n\\25.47/2.83 = [0.819/0.273 m]^x\\9 = 3^x\\x = 2[/tex]

Now it will be possible to find the value of the constant as:

[tex]2.83 = k [0.273 m]^2 [0.763]^0\\k = 2.83/0.0745\\k = 37.98 m^{-1}s^{-1}[/tex]

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