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adelina 88 [10]
2 years ago
10

Somebody please help me thank

Mathematics
2 answers:
stira [4]2 years ago
8 0

Answer:

Step-by-step explanation:

5.003*10^{9}=5,003,000,000

5.003* 10⁹ < 5,083,000,000

zepelin [54]2 years ago
6 0
5.003 x10^9 < 5,083,000,000 x10^-1
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Find the components of the vector v with initial point P and terminal point Q. Find the unit vector u in the direction of v.
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Answer:

Step-by-step explanation:

v=Q-P

v=(5,1,0)

u=v/(magnitude of v)=(5,1,0)/(\sqrt{5^2+1^2+0^2)

u=(5/5.09,1/5.09,0)

v=Q-P

v=(1,1,0)

u=v/(magnitude of v)=(1,1,0)/(\sqrt{1^2+1^2+0^2)

u=(1/\sqrt{2},1/\sqrt{2},0)

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2 years ago
A relay race has 4 runners who run different parts of the race. There are 16
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Your coach can select it 4 times I think
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Luis wants to buy a skateboard that usually sells for $79.08. All merchandise is discounted by 12%. What is the total cost of th
FrozenT [24]

Answer:

$74.96

Step-by-step explanation:

$79.80 × 0.12 = 9.576 (discount)

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$70.224 + $4.74012 = $74.96412

8 0
3 years ago
At 2:00pm a car's speedometer reads 20mph, and at 2:10pm it reads 30mph.
Dmitry [639]

Over this 10-minute interval, the car's average acceleration is

\dfrac{30\,\mathrm{mph}-20\,\mathrm{mph}}{10\,\mathrm{min}}=\dfrac{10\,\mathrm{mph}}{\frac16\,\mathrm h}=60\dfrac{\rm mi}{\mathrm h^2}

The MVT says that at some point during this 10-minute interval, the car must have had an acceleration of 60 mi/h^2.

6 0
3 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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