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3 years ago

I WILL MARK YOU BRAINLIEST!!!!

Mathematics

1 answer:

Stella [2.4K]3 years ago

8 0

I’m pretty sure it’s option D

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-6-2b=-8. What's the answer

Ksivusya [100]

Answer: B = 1

Step-by-step explanation: The first step is to add 6 to both sides of the equation to get -2b = -2

You then divide both sides by negative 2 so you can get a positive

dividing both sides by negative 2 will leave you with 1b = 1

b = 1

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2 years ago

Find the exact value cos5pi/6

Lelechka [254]

Answer:

$- \frac{ \sqrt{3} }{2}$

Step-by-step explanation:

Unit circle

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3 years ago

Please help me with this questions

viktelen [127]

Answer:

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Step-by-step explanation:

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3 years ago

(25 points!) Caleb bought a $795 necklace for his wife's birthday. If he bought it on credit with an interest rate of 16.25% how

Sliva [168]

Answer:

The total amount that he would have paid after 3 years is $1183.

Step-by-step explanation:

We would apply the formula for determining simple interest which is expressed as

I = PRT/100

I represents interest paid on the amount owed.

P represents the principal or initial amount owed.

R represents interest rate

T represents the duration for which the amount was owed in years.

P = $795

R = 16.25

T = 3 years

I = (795 × 16.25 × 3) ÷ 100

I = $388

Total amount that he would have paid after 3 years is $795 + $388 = $1183

7 0

3 years ago

The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas

saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by $s(t) = -t^3+3t^2+7t +4$ , where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0

3 years ago

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