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3 years ago

2. Using 50 for the number of weeks in a year is a(n) exponent

Mathematics

1 answer:

Paladinen [302]3 years ago

6 0

There are 52 weeks in a year.

Using 50 for the number of weeks in a year is a(n) round.

Rounding is a process which we approximate the known number by just scaling it up or down.

In rounding the numbers less than 5 are scaled to one number less than the given integer and numbers such as 5 or above than 5 are scaled a number greater than the given number .

For example if there were 55 or more than weeks in a year we can say there are approximately 60 weeks in a year.

But there are 52 weeks so we round it to 50 weeks as 2 is less than 5.

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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

Read 2 more answers

Marcus can walk 1 mile to the library in the same time he can bicycle 2.5 miles to the sporting goods store. His speed on the bi

Maksim231197 [3]

Let
x------> speed on the bicycle
y------> speed walking

we know that
x=6+y------> equation 1
speed=distance /time-------> time=distance/speed
1/y=2.5/x-------> y=x/2.5------> x=2.5y-------> equation 2

equals 1 and 2
6+y=2.5y-----> 2.5y-y=6------> 1.5y=6-----> y=6/1.5-----> y=4 mi/h
x=6+y-----> x=6+4-----> x=10 mi/h

the answer is
the speed walking is 4 mi/h

6 0

4 years ago

Malika decides to use some of her savings to buy a bicycle. She compares the prices of 3 brands of bicycles: the RacerZ, the Spe

jeka94

So times $64.80 by 2.5. You minus 20 from that, so that the RacerZ is $142. You take that $142 and times that by 1.09. You get 154.78. Therefore;

Speedmonster- 64.80
RacerZ- 142
Swiftee-154.78

Hope this helps. :)

3 0

3 years ago

let sin(θ) =3/5 and tan(y) =12/5 both angels comes from 2 different right trianglesa)find the third side of the two tringles b)

statuscvo [17]

In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

$\sin (\theta)=\frac{3}{5}$

Describes the following triangle

To find the missing length x, we could use the Pythagorean Theorem. The sum of the squares of the legs is equal to the square of the hypotenuse. From this, we have the following equation

$x^2+3^2=5^2$

Solving for x, we have

$\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}$

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

$\tan (y)=\frac{12}{5}$

Describes the following triangle

Using the Pythagorean Theorem again, we have

$5^2+12^2=h^2$

Solving for h, we have

$\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}$

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

$\begin{gathered} \sin (\theta)=\frac{3}{5} \\ \cos (\theta)=\frac{4}{5} \\ \sin (y)=\frac{12}{13} \\ \cos (y)=\frac{5}{13} \end{gathered}$

To calculate the sine and cosine of the sum

$\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}$

We can use the following identities

$\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}$

Using those identities in our problem, we're going to have

$\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}$

4 0

1 year ago

Solve for y. show work. answer may be left as decimal. 0.4(y+10)+0.6y=2

maxonik [38]

Answer: y=2

Step-by-step explanation:

combine like terms 2+ -4 = .2 1y= -2

divide each side my 1 then simplifying y=2

4 0

3 years ago

Read 2 more answers