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GrogVix [38]
3 years ago
14

Please help im desperate no. 48 and 49​

Mathematics
2 answers:
stealth61 [152]3 years ago
8 0

Answer:

C and (3, 6 )

Step-by-step explanation:

(48)

Since given y = 2x the best method of solving is by substitution C

(49)

Given the 2 equations

3x + y = 15 → (1)

y = 2x → (2)

Substitute y = 2x into (1)

3x + 2x = 15

5x = 15 ( divide both sides by 5 )

x = 3

Substitute x = 3 into (2) and evaluate for y

y = 2x = 2 × 3 = 6

solution is (3, 6 )

exis [7]3 years ago
3 0

Answer:

48. option C : Substitution

49. Option D :  ( 3, 6)

Step-by-step explanation:

3x + y = 15       ------------- ( 1 )

      y = 2x  ------------------- ( 2 )

Substitute ( 2 ) in ( 1 ) :

3x + y = 15

3x + 2x = 15

5x = 15

x = \frac{15}{5} = 3

Substitute x in ( 2 ) :

y = 2x

y = 2 × 3 = 6

The solution to the system of equation is ( 3 , 6 )  

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alina1380 [7]

Answer:

\displaystyle{4\tan x + \sin 2x - 6x + C}

Step-by-step explanation:

We are given the integral of:

\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}

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\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}

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\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}

Since \displaystyle{\sec x = \dfrac{1}{\cos x}} then it can be simplified to:

\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}

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\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}

For \displaystyle{\cos ^2 x}, we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

\displaystyle{2\cos^2 x -1 = \cos2x}

We can solve for \displaystyle{\cos ^2x} which is:

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\displaystyle{4\tan x + \sin 2x - 6x + C}

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6
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