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attashe74 [19]
3 years ago
9

Mai spends 7 and 3/5 hours in school each day. Her lunch period is 30 minutes long, and she spends a total of 42 minutes switchi

ng rooms between classes. The rest of her day is spent in 6 classes that are all the same length. How long is each class?
Mathematics
1 answer:
Sonja [21]3 years ago
8 0
First, we solve for the number of minutes in 7 and 3/5 hours by multiplying the number by 60 giving us,
                            (7 + 3/5) x (60) = 456 minutes
Spending 30 minutes for lunch will leave her with 426 minutes. Then, spending 42 minutes for switching of classes will finally give her 414 minutes. 

We then divide this value by 6 (for her 6 classes) giving us 69 minutes. Thus, each class is 69 minutes long. 
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In a dilation with a scale factor of 3/4,the image is __________ the preimage
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sergey [27]
These are two questions and two answers.

1) Problem 17.

(i) Determine whether T is continuous at 6061.

For that  you have to compute the value of T at 6061 and the lateral limits of T when x approaches 6061.

a) T(x) = 0.10x if 0 < x ≤ 6061

T (6061) = 0.10(6061) = 606.1

b) limit of Tx when x → 6061.

By the left the limit is the same value of T(x) calculated above.

By the right the limit is calculated using the definition of the function for the next stage: T(x) = 606.10 + 0.18 (x - 6061)

⇒ Limit of T(x) when x → 6061 from the right = 606.10 + 0.18 (6061 - 6061) = 606.10

Since both limits and the value of the function are the same, T is continuous at 6061.

(ii) Determine whether T is continuous at 32,473.

Same procedure.

a) Value at 32,473

T(32,473) = 606.10 + 0.18 (32,473 - 6061) = 5,360.26

b) Limit of T(x) when x → 32,473 from the right

Limit = 5360.26 + 0.26(x - 32,473) = 5360.26

Again, since the two limits and the value of the function have the same value the function is continuos at the x = 32,473.

(iii) If  T had discontinuities, a tax payer that earns an amount very close to the discontinuity can easily approach its incomes to take andvantage of the part that results in lower tax.

2) Problem 18.

a) Statement Sk

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Sk = 1 + 4 + 7 + ... (3k - 2) = k(3k - 1) / 2

b) Statement S (k+1)

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Simplification:

1 + 4 + 7 + ... + 3k - 2+ 3k + 3 - 2] = (k + 1) (3k + 3 - 1)/2

                 k(3k - 1)/ 2 + (3k + 1) = (k + 1)(3k+2) / 2

Do the operations on the left side and  you will find it can be simplified to k ( 3k +1) (3 k + 2) / 2.

With that you find that the left side equals the right side which is a proof of the validity of the statement by induction.

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