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andrew-mc [135]
3 years ago
7

If 2/3 is multiplied by a fraction greater than 1, which of these could be the product?

Mathematics
1 answer:
nadezda [96]3 years ago
3 0

Answer:

Any number bigger than 2/3

Step-by-step explanation:

btw, I didn't see the options but here's the explanation

When you multiply by a part of a whole, a fraction, you are getting smaller. When you multiply by a number bigger than one, your product gets bigger. So, any number bigger than 2/3 works just fine

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A closet is 2 feet deep , 5 feet wide , and 7 feet tall what is the valume of the closet in cubic feet , rounded to the nearest
slavikrds [6]

Answer:

70 ft

Step-by-step explanation:

2x5x7=70

5 0
3 years ago
A cosine function is graphed below. Use the drop-down menus to describe the graph. The amplitude of the graph is __ . The equati
Tresset [83]

Amplitude:4

Equation of Midline: 2

Period of function:3

Function shifted left:0.5

Function shifted up: 2

4 0
3 years ago
Read 2 more answers
1. Derive the half-angle formulas from the double
lilavasa [31]

1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following <em>trigonometric</em> formulae are used to derive the <em>half-angle</em> formulas:

sin² θ / 2 + cos² θ / 2 = 1                      (1)

cos θ = cos² (θ / 2) - sin² (θ / 2)           (2)

First, we derive the formula for the sine of a <em>half</em> angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a <em>half</em> angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a <em>half</em> angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in <em>rectangular</em> form to <em>polar</em> form are obtained by <em>trigonometric</em> functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in <em>polar</em> form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in <em>rectangular</em> form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the <em>linear</em> function y = 5 · x - 8, we proceed to use the following <em>substitution</em> formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

#SPJ1

4 0
2 years ago
Solve for x.<br> 2x + 4<br> 4x - 88<br> x = [?]
baherus [9]
Solve for x.
2x + 4= 8
4x - 88 = -352
x = [?]
7 0
3 years ago
A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.
mezya [45]
1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= \frac{1}{2}sin60^{o}side*side=   \frac{1}{2} \frac{ \sqrt{3} }{2}  (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}
A=\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2}  )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 2 \pi r=x

so r= \frac{x}{2 \pi }

Area of circle = \pi  r^{2}= \pi  ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2}  }* x^{2} = \frac{1}{4 \pi } x^{2}

4. Let f(x)=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0

(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b

x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

6. So one part is \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) } and the other part is b-\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }

4 0
3 years ago
Read 2 more answers
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