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statuscvo [17]
2 years ago
12

Find the angle in the attachment​

Mathematics
1 answer:
Vadim26 [7]2 years ago
7 0
Right-angled triangle. Let x be the required angle. Tan x = 8/6. x= 53.1°
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What are the points on the lines???
den301095 [7]

Answer:

The top is (-2 , 2)  and the bottom is (4 , -2)

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Translate this question into an equation What percent of 196 is 49?
Marizza181 [45]

Answer:

P * 196 = 49

Step-by-step explanation:

What percent of 196 is 49

Percent is P

of means multiply is means equals

P * 196 = 49

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Write the word namefor the decimal 31.903
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Thirty-one and 903 thousandths
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Show the work pleasee
Nataly [62]

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a

Step-by-step explanation:

6 0
3 years ago
2 more questions thanks
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

You just need to replace n for k:

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

b) Statement S (k+1)

Replace

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

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.

4 0
3 years ago
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