All categories

3 years ago

Jason writes an expression equivalent to -2 + 5(d + 1) - 6d. His work is shown below.

1 answer:

5 0

The step in which Jason made the error is; <em><u>the first step and the error he made was that he didn't obey the BODMAS rule of algebra operations</u></em>.

We are told that Jason wants to write an expression that is equivalent to -2 + 5(d + 1) - 6d.

Now, his steps are;

Step 1; 3(d + 1) - 6d

There is already an error here because 5 is attached to the bracket and not standing alone.

Therefore, Jason should have used distributive property to multiply 5 by the numbers inside the bracket instead of treating the 5 like it is isolated to be added to -3. This is because in BODMAS, Bracket comes before addition.

Read more at; brainly.com/question/14646272

You might be interested in

Answer with all calculations

Sloan [31]

Answer:

Part a) The acceleration in the first 20 seconds is $0.35\ m/sec^2$

Part b) The distance between the two stops is 945 meters

Part c) The average speed during the journey is 5.9 m/sec

Step-by-step explanation:

Part a)

Let

x ---> the time in seconds

y ---> the velocity in m/sec

a ---> acceleration in m/ sec^2

we know that

The acceleration is equal to divide the velocity by the time

The acceleration is the slope of the linear equation of the graph

For the first 20 seconds

take the points

(0,0) and (20,7)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute

$m=\frac{7-0}{20-0}$

$m=\frac{7}{20}=0.35\ m/sec^2$

therefore

The acceleration in the first 20 seconds is $0.35\ m/sec^2$

Part b) we know that

Looking at the graph

The two stops of the bus are the points (0,0) and (160,0)

The stops of the bus represent the x-intercepts of the linear equation

Values of x when the value of y is equal to zero

In the context of the problem

Values of the time when the velocity of the bus is equal to zero

To find out the distance, we need to determine the area of the graph

(because the distance is equal to multiply the velocity by the time)

The area of the graph in the interval [0,160] is equal to the area of two right triangles plus the area of rectangle

$A=\frac{1}{2}(20)( 7)+(130-20)(7)+ \frac{1}{2}(160-130)(7)$

$A=70+770+105=945\ m$

Part c) we know that

The average speed is equal to divide the total distance by the total time

$\frac{945}{160}=5.9\ m/sec$

7 0

3 years ago

The sum of the first 65 odd, positive integers is?

Aleksandr [31]

$1+3+5+7+9+11+...\\\\This\ is\ an\ atihmetic\ progression\ where\ a_1=1\ and\ d=2.\\\\The\ sum:S_n=\frac{2a_1+(n-1)d}{2}\cdot n\\\\S_{65}=\frac{2\cdot1+(65-1)\cdot2}{2}\cdot65=\frac{2+64\cdot2}{2}\cdot65=\frac{2+128}{2}\cdot65\\\\=\frac{130}{2}\cdot65=65\cdot65=4225$

7 0

4 years ago

4. [5 pts] Describe how and why the formula for permutations differs from the formula for combinations.

ivann1987 [24]

Answer:

They are different because of the order in the permutation matters. In combination, the order doesn't matter. In other words in a permutation 123 and 132 are different but in a combination are the same group (they have the same digits 1,2, and 3).

Step-by-step explanation:

The formula of the permutation is $P(n,r)=\frac{n!}{(n-r)!}$ , when you are performing a permutation you pick r objects from a total of n, for the first pick you can choose from n, but for the second you have n-1, and this continues to your pick number r in which you will choose from n-r+1, and the total of permutation is the multiplicación of this number of choices for each pick, like this:

$n(n-1)(n-2)...(n-r+1)$

If $n!=n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)$ and $(n-r)!=(n-r)(n-r-1)(n-r-2)...(1)$

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}$

The factor equals above and under cancel each other.

$\frac{n!}{(n-r)!}=\frac{n(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)...(1)}{(n-r)(n-r-1)(n-r-2)...(1)}\\\frac{n!}{(n-r)!}=n(n-1)(n-2)...(n-r+1)$

In combination, the order of the element isn't important, so from the total of permutation you have to eliminate the ones with the same objects with different order and counting just once each group, when choosing r objects the total of permutation for a single group of r objects is: $r(r-1)(r-2)...(1)=r!$ . If you divide the total of permutations of n taking r by r! you get the combinations (where the order is not important). The formula of the combination is $C(n,r)=\frac{n!}{r!(n-r)!}$ .