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3 years ago

Bobcat Park is a rectangular park with an area of 5 2/5 square miles.Its width is 1 11/25 miles.How long is the park?

Mathematics

1 answer:

ELEN [110]3 years ago

4 0

I believe it is 3.75 miles. 5 2/5=5.4 and 1 11/25=1.44. 5.4/1.44=3.75.

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35° 50° ? Find the measure of the missing angle:

Marianna [84]

Answer:

i think the answer would be 95 degree. im not absolutley positive though

Step-by-step explanation:

i added 35 and 50 and got the sum of 85 then i subtracted 85 from 180 and got 95

feedback is appreciated.

(i'm not gonna beg for brainliest)

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3 years ago

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

bearhunter [10]

Answer: See Below

Step-by-step explanation:

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 $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

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

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

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

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

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

Hint: sin and cos are only opposite signs in Quadrants II and IV

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

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In degrees: t = 120° + 360n and 300° + 360n

4 0

3 years ago

The revenue from selling xshirts is r(x) = 12x.

algol [13]

Answer:

D. $p(x)=7x-20$

Step-by-step explanation:

$r(x) = 12x\\c(x) = 5x + 20\\p(x) = r(x)-c(x)\\p(x)=12x-(5x+20)\\p(x)=12x-5x-20\\p(x)=7x-20$

Simply subtract the two functions from each other and simplify to get the result.

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3 years ago

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gayaneshka [121]

Answer:

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Step-by-step explanation:

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3 years ago

WE has endpoints at W(10, 2) and E(5, -3). How long is WE?

s2008m [1.1K]

Answer:

$WE=5\sqrt{2}$

Step-by-step explanation:

Distance between two points:

The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is given by $=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2$

Hence distance between $W(10,2)$ and $E(5,-3)$

$WE=\sqrt{(5-10)^2+(-3-2)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}\\\\WE=\sqrt{50}=5\sqrt{2}$

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3 years ago