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3 years ago

Solve (2 - x)^2 < 4/25

Mathematics

2 answers:

uysha [10]3 years ago

6 0

Answer:

x>8/5

Step-by-step explanation:

(2-x)^2<4/25

take the square root of both sides

Semenov [28]3 years ago

5 0

Answer:

8/5 < x< 12/5

Step-by-step explanation:

(2 - x)^2 < 4/25

Take the square root of each side

sqrt((2 - x)^2) <±sqrt( 4/25)

Make two equations

2-x < 2/5 2-x > -2/5

Subtract 2 from each side

2-x-2 < 2/5 -2 2-x-2 > -2/5-2

-x < 2/5 - 10/5 -x > -2/5 - 10/5

-x < -8/5 -x > -12/5

Multiply by -1, remembering to flip the inequality

x> 8/5 x < 12/5

8/5 < x< 12/5

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Points F(3, 3), K(6, 5), J(8, 2) and L(5, 0) lie in a coordinate plane. Which lines are perpendicular?

Romashka-Z-Leto [24]

To determine if a line is perpendicular to another, you must first determine the slope...
m = y1-y2/x1-x2
m of FK = 3-5/3-6 = -2/-3 = 2/3
m of FJ = 3-2/3-8 = 1/-5
m of FL = 3-0/3-5 = 3/-2
m of KJ = 5-2/6-8 = 3/-2
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In order for two lines to be perpendicular, their slopes must be opposite reciprocals...
FK is perpendicular to FL
FK is perpendicular to KJ
JL is perpendicular to FL
JL is perpendicular to KJ
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8 0

3 years ago

Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end frac

uranmaximum [27]

One way to do this is to notice

$\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4$

Then

$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$ ,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$

7 0

3 years ago

Graph {x | x > 4} {x | x < -2} on a number line

WARRIOR [948]

Answer:

Step-by-step explanation:

well,

you mark an arrow to the left side of # line with open circle on -2

then,

you mark an arrow to the right side of # line with open circle on 4

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3 years ago

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