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Rina8888 [55]
3 years ago
13

The Heat (H) required to melt copper is proportional to the copper’s mass in grams (g). Suppose the number of calories to melt 1

gram of copper is represented by c . Which statements are true for this situation

Mathematics
1 answer:
Aleks [24]3 years ago
7 0

Answer:

H = cg The number of calories to melt one gram is the constant of proportionality. Heat = calories(mass) Write as y = kx, where k is the constant of proportionality → H = cg → c is the constant of proportionality

Step-by-step explanation:

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Rosa wants 2 pairs of jeans that cost $32 each .she has a coupon for $10 off the price of one pair .what price will rosa pay for
Phoenix [80]

Answer:

$54

Step-by-step explanation:

An expression for the total cost would be:

1($32) + ($32-$10) = $32 + $22 = $54

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a rectangular lawn of sides 400m and 300m is surrounded by a path of width 3 m . find the area of path
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2109 meter squared

Step-by-step explanation:

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In Exercises 5-7, find all the exact t-values for which the given statement is true,
bearhunter [10]

Answer:  See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians:     t = 0π + 2πn

In degrees:   t = 0° + 360n

******************************************************************************

6)\quad sin (t) = \dfrac{1}{2}

Where on the Unit Circle does   sin = \dfrac{1}{2}

<em>Hint: sin is only positive in Quadrants I and II</em>

\text{Answer: at}\  \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n

In degrees:    t = 30° + 360n  and  150° + 360n

******************************************************************************

7)\quad tan (t) = -\sqrt3

Where on the Unit Circle does    \dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

\text{Answer: at}\  \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)

\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n

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