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3 years ago

14

Use trigonometric expressions to build an equivalent trigonometric identity with the given expression: sec (x) tan (x) cos (x) c

sc (x) =

2 answers:

Bess [88]3 years ago

8 0

Answer:

sec (x)

Step-by-step explanation:

sec (x) tan (x) cos (x) csc (x) =

We know sec = 1/ cos

Tan = sin/cos

csc = 1/sin

Replacing into the expression

1/ cos (x) * sin(x)/ cos (x) * cos (x) * 1 / sin(x)

Canceling like terms

1/ cos (x)

sec(x)

alexira [117]3 years ago

8 0

Step-by-step explanation:

<u>Step 1: Simplify all of the trigonometric functions</u>

$sec(x) = \frac{1}{cos(x)}$

$tan(x) = \frac{sin(x)}{cos(x)}$

$cos(x) = cos(x)$

$csc(x) = \frac{1}{sin(x)}$

$\frac{1}{cos(x)}*\frac{sin(x)}{cos(x)}*\frac{cos(x)}{1}*\frac{1}{sin(x)}$

$\frac{sin(x)cos(x)}{cos(x)cos(x)sin(x)}$

$\frac{1}{cos(x)}$

$sec(x)$

Answer: $sec(x)$

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

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Solving the expression

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$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

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$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

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3 years ago

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Afina-wow [57]

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2 years ago

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Find the coordinates of the point (x,y,z) on the planez=4x+3y+1 which is closest to the origin.

Nataliya [291]

Given plane Π : f(x,y,z) = 4x+3y-z = -1
Need to find point P on Π that is closest to the origin O=(0,0,0).

Solution:
First step: check if O is on the plane Π : f(0,0,0)=0 ≠ -1 => O is not on Π
Next:
We know that the required point must lie on the normal vector <4,3,-1> passing through the origin, i.e.
P=(0,0,0)+k<4,3,-1> = (4k,3k,-k)
For P to lie on plane Π , it must satisfy
4(4k)+3(3k)-(-k)=-1
Solving for k
k=-1/26
=>
Point P is (4k,3k,-k) = (-4/26, -3/26, 1/26) = (-2/13, -3/26, 1/26)
because P is on the normal vector originating from the origin, and it satisfies the equation of plane Π
Answer: P(-2/13, -3/26, 1/26) is the point on Π closest to the origin.

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