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3 years ago

Which one is it? will mark brainliest only if correct

Mathematics

2 answers:

nata0808 [166]3 years ago

6 0

Answer:

D, 1080

Step-by-step explanation:

degrees = (n-2)180, if n is the number of sides.

Sonja [21]3 years ago

4 0

Answer:

540°

Oct = 1080 °

Cir = 2160 °

2160 ° ÷ 1080 ° = 2

1080 ° ÷ 2 = 540°

Step-by-step explanation:

Mark as brainllest

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9282737 + 1881 x 19916 =

sweet-ann [11.9K]

Answer to this math equation: 46744733

5 0

3 years ago

Read 2 more answers

5.03+13.7+108 what is the answer an how do u solve it

jarptica [38.1K]

Answer:

126.73

Step-by-step explanation:

5.03+13.7+108

5.03+13.7= 18.73

18.73 +108=

126.73

5 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

A student has a savings account earning 3% simple interest. She must pay $1200 for first-semester tuition by September 1 and $12

Bingel [31]

Using simple interest, it is found that she needs to earn $2,391.07 during the summer.

<h3>Simple Interest</h3>

Simple interest is used when there is a single compounding per time period.

The amount of money after t years in is modeled by:

A(t) = A(0)(1 + rt)

In which:

A(0) is the initial amount.

r is the interest rate, as a decimal.

For this problem, the objective is to have <u>$1200 in 3 months = 0.25 years</u>, hence the parameters are given as follows:

A(0.25) = 1200, t = 0.25, r = 0.03.

Hence we have to solve for A(0):

A(0)(1 + 0.03 x 0.25) = 1200

A(0) = 1200/(1 + 0.03 x 0.25)

A(0) = $1,191.07.

She also needs to earn $1,200 to pay the first-semester bill on time, hence:

1200 + 1191.07 = $2,391.07.

She needs to earn $2,391.07 during the summer.

More can be learned about simple interest at brainly.com/question/16646150

#SPJ1

3 0

2 years ago

A small business makes 3-speed and 10-speed bicycles at two different factories. Factory A produces 16 3-speed and 20 10-speed b

GarryVolchara [31]

Answer:

Factory A should be operated 6.11 days

Factory B should be operated 6.88 days

The minimum cost for factory A is $6,110

The minimum cost for factory B is $5,504

Step-by-step explanation:

Factory A

Daily production: 16 3-speed and 20 10-speed bikes

Total daily production = 16+20 = 36 speed bikes

Order: 80 3-speed and 140 10-speed bikes

Total order = 80+140 = 220 speed bikes

Number of days = 220/36 = 6.11 days

Cost per day = $1,000

Minimum cost for 6.11 days = $1000 × 6.11 = $6,110

Factory B

Daily production: 12 3-speed and 20 10-speed bikes

Total daily production = 12+20 = 32 speed bikes

Total order = 220 speed bikes

Number of days = 220/36 = 6.88 days

Cost per day = $800

Cost for 6.88 days = $800 × 6.88 = $5,504

6 0

3 years ago