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Fantom [35]
3 years ago
15

Write 6208 in exponential form​

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

2^6 x 97

Step-by-step explanation:

In order to factor an integer, we need to repeatedly divide it by the ascending sequence of primes (2, 3, 5...).

The number of times that each prime divides the original integer becomes its exponent in the final result.

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(\tan ^(2)\theta \cos ^(2)\theta -1)/(1+\cos (2\theta ))=
Vitek1552 [10]

(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

Recall that

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:

(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))

Recall the double angle identity for cosine,

cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

so the 1 in the denominator also vanishes:

(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))

Recall the Pythagorean identity,

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

which means

sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):

-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))

Cancel the cos²(<em>θ</em>) terms to end up with

(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2

7 0
2 years ago
0.6 as a fraction and percent
Ludmilka [50]
6/10 is the fraction and 60 percent
6 0
3 years ago
Read 2 more answers
Mary looks over reports on four of her workers. jack made 30 baskets in 5 hours. rudy made 32 baskets in 8 hours. sam made 40 ba
Elena L [17]
Well let's do some maths if each made a certain amount of baskets an hour you could divide them and figure out how many baskets were made per hour.
jack = 30/5 = 6 baskets per hour
Rudy = 32/8 = 4 baskets per hour
Sam = 40/12 =3.333 baskets per hour
Walter = 22/4 = 5.5 baskets per hour
according to this Jack had the greatest productivity
6 0
3 years ago
Eight sided dice can be described as two square based pyramids stuck together at
kobusy [5.1K]
It costs $3.63 more for the gold die than the gold plated die.

The volume of each half of the die is given by the formula

V=1/3s³.

This means the total volume is V = 2(1/3s³) = 2(1/3(2³)) = 16/3 = 5.33 mm³.

For the solid gold die, we know that it weighs 0.02g per mm³:

5.33(0.02) = 0.1066g

Gold costs $40.38 per gram:
40.38(0.1066) = $4.30

For the plastic die, plastic weighs 0.01 g per mm³:
0.01(5.33) = 0.0533g

Plastic costs $10 per gram:
0.0533(10) = 0.533 ≈ $0.54.

We also need to find the surface area, as it will be painted.  Each face is an equilateral triangle; the area of a triangle is given by the formula

A = 1/2bh

We need to use the Pythagorean theorem to find the height.  The height will be one leg and 1/2 of the base will be the other; the slanted side will be the hypotenuse:

h² + (1/2(2))² = 2²
h² + 1² = 2²
h² + 1 = 4
h² = 3
h = √3

The area of the triangle would then be A = 1/2bh = 1/2(2)(√3) = √3.  There are 8 such faces on the die, so the total surface area would be 8√3 = 13.86mm².

Paint costs $0.01 per mm²:
13.86(0.01) = 0.1386 ≈ $0.14

Together the cost for the plastic is 0.14+0.53 = $0.67.

The difference between it and the cost for the gold die is 4.30 - 0.67 = $3.63.
5 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}

F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
2 years ago
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