Mr. Lewis sees a red light ahead and applies the brakes on his car. The car travels 300.5 feet before coming to a stop. For each foot the car travels during this time, its speed changes by -0.2 miles per hour. What is the total change in the car's speed during that time? ( PLEASE ANSWER AS SOON AS POSSIBLE )

A. -36.5
B. -60.1
C. -60.10
D. 36.5

The loss of each owner of the catering company is 22 dollars.

Average is the sum of the total weights of each individual divided by the number of individuals.

Given that a catering company loses 110 dollars as a result of the delay and the company has 5 owners and they share the loss equally.

As the loss has been shared by the owners equally we have to divide the total loss occurred by the total no. of owners to obtain the amount of loss for each individual which is,

= 110/5 dollars.

= 22 dollars.

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On simplification of inequality x + 10 > 2, we get x>-8.

The given question is related to algebra and inequalities.

To solve the inequality 'x + 10 > 2', we need to isolate x.

Subtract 10 from both sides: x + 10 - 10 > 2 - 10.

This simplifies to: x > -8.

The solution of the equation 14 = 2c − 6 + 3c will be 4.

The distribution of weights to the variables involved that establishes the equilibrium in the calculation is referred to as a result.

Making anything easier to accomplish or comprehend, as well as making it less difficult, is the definition of simplification.

The equation is given below.

14 = 2c − 6 + 3c

Simplify the equation, then the value of the variable c will be

14 = 2c − 6 + 3c

20 = 5c

c = 20 / 5

c = 4

The solution of the equation 14 = 2c − 6 + 3c will be 4.

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Answer: 24 Units squared

Step-by-step explanation:

Answer:

The given graph is : f(x) = [tex]3^x[/tex]

The  graph g(x) is transformation of graph f(x).

The curve g(x) passes through (-2,2) and (-1,4).

Let (a,b) be any point on f(x).After translation  new curve g(x) is obtained which is parallel to line, y=1 or asymptode is ,y=1 or when y becomes 1 the graph becomes parallel to x axis.

g(x)= y-1=[tex]3^{x+2}[/tex]

g(x) = y = [tex]3^{x+2} +1[/tex] is the graph of translated curve.

The two graph is depicted below.

To add fractions 1/9 and 5/6, find the least common denominator, convert each fraction to have this denominator, then add the numerators while keeping the denominator the same. The resulting fraction is 17/18.

The first thing you need to do to add fractions 1/9 and 5/6 is to find a common denominator. Since we can only add numerators with the same denominator, we must convert these fractions so that they have the same bottom number. The least common denominator (LCD) for 9 and 6 is 18.

So, we convert 1/9 by multiplying both its numerator and denominator by 2, to get 2/18. Similarly, we convert 5/6 by multiplying both its numerator and denominator by 3, to get 15/18. Now we can add the numerators:

2/18 + 15/18 = 17/18

Remember, we never add the denominators; they remain the same. After finding a common denominator and converting each fraction, we add the numerators together to get the final result.

Final answer:

To add 1/9 and 5/6, find a common denominator, convert each fraction to an equivalent fraction with this denominator, and then add the numerators. The least common multiple of 9 and 6 is 18, leading to 1/9 + 5/6 = 17/18 after conversion and addition.

Explanation:

The first thing you need to do to add the fractions 1/9 and 5/6 is to find a common denominator. A common denominator is necessary because you can only add fractions directly when they share the same denominator. To find a common denominator, you can often multiply the denominators together, but it's better to find the least common multiple (LCM) of the two denominators to simplify the process.

Once the common denominator is found, you convert each fraction to an equivalent fraction with this common denominator. Then, you add the numerators while keeping the denominator the same. The denominators are never added together. If the result is an improper fraction, you may have to simplify it or convert it to a mixed number.
Using our given fractions, the LCM of 9 and 6 is 18. Therefore, 1/9 can be converted to 2/18 and 5/6 to 15/18. Adding these gives 2/18 + 15/18 = 17/18.

Total time driving to and from beach: 12.3 hours; averaging 58 mph there, 65 mph returning home.

To find the total time Paul's family spent driving to and from the beach, we can calculate the time for each leg of the trip and then add them together.

1. Time taken for the trip to the beach:

[tex]\[ \text{Time}_1 = \frac{\text{Distance}_1}{\text{Speed}_1} \][/tex]

2. Time taken for the return trip:

[tex]\[ \text{Time}_2 = \frac{\text{Distance}_2}{\text{Speed}_2} \][/tex]

Given:

[tex]- Distance to the beach, \( \text{Distance}_1 = 377 \) miles\\- Speed to the beach, \( \text{Speed}_1 = 58 \) mph\\- Speed returning home, \( \text{Speed}_2 = 65 \) mph[/tex]

Let's calculate:

[tex]1. \[ \text{Time}_1 = \frac{377}{58} \][/tex]

[tex]2. \[ \text{Time}_2 = \frac{377}{65} \][/tex]

Now, let's compute:

[tex]\[ \text{Time}_1 = \frac{377}{58} \approx 6.5 \text{ hours} \]\[ \text{Time}_2 = \frac{377}{65} \approx 5.8 \text{ hours} \][/tex]

Finally, to find the total time spent driving to and from the beach, we add these times:

[tex]\[ \text{Total time} = \text{Time}_1 + \text{Time}_2 \]\[ \text{Total time} \approx 6.5 + 5.8 \]\[ \text{Total time} \approx 12.3 \text{ hours} \][/tex]

So, Paul's family spent approximately 12.3 hours driving to and from the beach.

Three of the numbers are correctly written in scientific notation. For the two that are not, we rewrite them correctly.

A. 4.51x10-2: This number is correctly written in scientific notation.

B. 0.789x105: This number is not correctly written in scientific notation. To rewrite it correctly, we move the decimal point to the right so that there is only one non-zero digit to the left of the decimal point. This gives us 7.89x104.

C. 31.5x102: This number is correctly written in scientific notation.

D. 3.008x10-8: This number is not correctly written in scientific notation. To rewrite it correctly, we move the decimal point to the left so that there is only one non-zero digit to the left of the decimal point. This gives us 3.008x10-9.

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Answer:

[tex]2 a^3 b^4 (a^3 - 11 a b + 8 b^3)[/tex]

Step-by-step explanation:

The given expression is

[tex]16a^3b^7 + 2a^6b^4 -22a^4b^5[/tex]

We need to find the factored form of the given expression.

In the given expression [tex]2 a^3 b^4[/tex] is the greatest common factor of all terms.

Taking GCF we get.

[tex]2 a^3 b^4 (8 b^3+a^3 - 11 a b)[/tex]

In can be written as

[tex]2 a^3 b^4 (a^3 - 11 a b + 8 b^3)[/tex]

This expression is the complete factored form of the given expression because [tex]a^3 - 11 a b + 8 b^3[/tex] can not be factored further.

Therefore, the factor form of given expression is [tex]2 a^3 b^4 (a^3 - 11 a b + 8 b^3)[/tex].

Answer: B) 5, −3, 1 over 2 are the zeros of f(x).

Step-by-step explanation:

Given function f(x)=[tex]2x^3-5x^2-28x+15[/tex]

If a is a zero of f(x) then f(a)=0

Let's check all the options

[tex]f(5)=2(5)^3-5(5)^2-28(5)+15\\=2(125)-5(25)-28(5)+15=250-125-140+15=0[/tex]

[tex]f(-3)=2(-3)^3-5(-3)^2-28(-3)+15\\=2(-27)-5(9)-28(-3)+15=-54-45+84+15=0[/tex]

[tex]f(\frac{-1}{2})=2\frac{-1}{2}()^3-5(\frac{-1}{2})^2-28(\frac{-1}{2})+15\\=2(\frac{-1}{8})-5(\frac{-1}{4})+14+15=\frac{-1}{4}-\frac{5}{4}+29=\frac{110}{4}=\frac{55}{2}\neq 0[/tex]

Therefore , 5 and -3 are zeroes of f(x).but -1/2 is not.

[tex]f(\frac{1}{2})=2\frac{1}{2}()^3-5(\frac{1}{2})^2-28(\frac{1}{2})+15\\=2(\frac{1}{8})-5(\frac{1}{4})-14+15=\frac{1}{4}-\frac{5}{4}-14+15=0[/tex]

⇒ 1/2 isthe third zero of f(x).

Here we got our three zeroes as 5,-3,1/2,rest other will not satisfy f(a)=0

Therefore B represents the zeroes of f(x).

The expression to represent the temperature 'h' hours after 1 p.m. given an initial temperature of 73°F and a decrease rate of 1.5°F per hour is 73 - 1.5h. This allows us to calculate the temperature at any given hour where 'h' is the number of hours after 1 p.m.

The problem involves dealing with linear equations, specifically those representing changes over time. In this situation, the temperature starts at 73°F at beginning of the time, i.e., at 1 pm and then decreases at a steady rate of 1.5°F per hour.

So, the decrease in temperature can be represented by 1.5h, where h is the number of hours elapsed since 1 p.m. As the temperature starts at 73°f, we subtract this decrease from the initial temperature. Therefore, the expression to determine the temperature after h hours is 73 - 1.5h.

This equation tells us that for each hour after 1 p.m., we simply subtract 1.5 from 73 and this would give the temperature at that time. For example, at 2 p.m., h equals 1, and the equation 73 - 1.5(1) gives us 71.5°F.

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Answer:

1. B. (8 x 20) - (8 x 1)

2. D. (2 x 9) + (2 x 3)

3. A. (7 x 50) - (7 x 2) = 350 - 14 = 336

4. C. (9 x 80) + (9 x 1) = 729