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Dmitry_Shevchenko [17]
3 years ago
14

I am trying to figure this question: 1 + tan^2A = Sec^2A

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

see explanation

Step-by-step explanation:

Using the Pythagorean identity

cos²A + sin²A = 1 ( divide terms by cos²A )

\frac{cos^2A}{cos^2A} + \frac{sin^2A}{cos^2A} = \frac{1}{cos^2A} , that is

1 + tan²A = sec²A ← as required

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15 points. Please explain your answer thanks so much.:)
Anni [7]

Answer:

CDJ= 34

Step-by-step explanation:

We know that ADJ = 90 degrees  

ADK = 180 degrees  and JDK equals 90 degrees so ADJ must equal 90 degrees

ADJ = ADC+CDJ

ADJ =4x+ 3x-8

Since ADJ = 90 degrees

90 = 4x+3x-8

Combine like terms

90 = 7x-8

Add 8 to each side

90+8 = 7x-8+8

98 = 7x

Divide each side by 7

98/7 = 7x/7

14 =x

We want to know CDJ

CDJ = 3x-8

CDJ = 3(14) -8

CDJ = 42-8

CDJ= 34

7 0
4 years ago
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Leona [35]

Answer: 54.4

Step-by-step explanation:

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3 years ago
If two lines are perpendicular, what must be true about the slopes of the two lines. Select all that apply.
bazaltina [42]

Answer:

Concept: Linear Systems

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7 0
3 years ago
Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constan
BARSIC [14]

Answer:

125/6(In(x-25)) - 5/6(In(x+5))+C

Step-by-step explanation:

∫x2/x1−20x2−125dx

Should be

∫x²/(x²−20x−125)dx

First of all let's factorize the denominator.

x²−20x−125= x²+5x-25x-125

x²−20x−125= x(x+5) -25(x+5)

x²−20x−125= (x-25)(x+5)

∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx

x²/(x²−20x−125) =x²/((x-25)(x+5))

x²/((x-25)(x+5))= a/(x-25) +b/(x+5)

x²/= a(x+5) + b(x-25)

Let x=25

625 = a30

a= 625/30

a= 125/6

Let x= -5

25 = -30b

b= 25/-30

b= -5/6

x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)

∫x²/(x²−20x−125)dx

=∫125/6(x-25) -∫5/6(x+5) Dx

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4 years ago
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