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podryga [215]
3 years ago
6

Pls answer you dont have to show work

Mathematics
1 answer:
Vlada [557]3 years ago
8 0

Answer:

Option C, 6.5

Step-by-step explanation:

<u>Step 1:  Substitute y in the equation 3x - 5y = 22</u>

3x - 5y = 22

3x - 5(-5x + 32) = 22

3x + 25x - 160 + 160 = 22 + 160

28x / 28 = 182 / 28

x = 6.5

Answer:  Option C, 6.5

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Answer:

See Explanation

Step-by-step explanation:

<em>Question like this are better answered if there are list of options; However, I'll simplify as far as the expression can be simplified</em>

Given

sec^4 x + sec^2 x tan^2 x - 2 tan^4 x

Required

Simplify

(sec^2 x)^2 + sec^2 x tan^2 x - 2 (tan^2 x)^2

Represent sec^2x with a

Represent tan^2x with b

The expression becomes

a^2 + ab- 2 b^2

Factorize

a^2 + 2ab -ab- 2 b^2

a(a + 2b) -b(a+ 2 b)

(a -b) (a+ 2 b)

Recall that

a = sec^2x

b = tan^2x

The expression (a -b) (a+ 2 b) becomes

(sec^2x -tan^2x) (sec^2x+ 2 tan^2x)

..............................................................................................................................

In trigonometry

sec^2x =1  +tan^2x

Subtract tan^2x from both sides

sec^2x - tan^2x =1  +tan^2x - tan^2x

sec^2x - tan^2x =1

..............................................................................................................................

Substitute 1 for sec^2x - tan^2x in (sec^2x -tan^2x) (sec^2x+ 2 tan^2x)

(1) (sec^2x+ 2 tan^2x)

Open Bracket

sec^2x+ 2 tan^2x ------------------This is an equivalence

(secx)^2+ 2 (tanx)^2

Solving further;

................................................................................................................................

In trigonometry

secx = \frac{1}{cosx}

tanx = \frac{sinx}{cosx}

Substitute the expressions for secx and tanx

................................................................................................................................

(secx)^2+ 2 (tanx)^2 becomes

(\frac{1}{cosx})^2+ 2 (\frac{sinx}{cosx})^2

Open bracket

\frac{1}{cos^2x}+ 2 (\frac{sin^2x}{cos^2x})

\frac{1}{cos^2x}+ \frac{2sin^2x}{cos^2x}

Add Fraction

\frac{1 + 2sin^2x}{cos^2x} ------------------------ This is another equivalence

................................................................................................................................

In trigonometry

sin^2x + cos^2x= 1

Make sin^2x the subject of formula

sin^2x= 1  - cos^2x

................................................................................................................................

Substitute the expressions for 1  - cos^2x for sin^2x

\frac{1 + 2(1  - cos^2x)}{cos^2x}

Open bracket

\frac{1 + 2  - 2cos^2x}{cos^2x}

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I got it right on the test

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