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2 years ago

Find the product of 3(-1/6) (-2/5)

Mathematics

1 answer:

Allisa [31]2 years ago

3 0

Answer:

= 1/5 ( 0.2 in decimal )

Step-by-step explanation:

3(-1/6) (-2/5)

= 1/5 ( 0.2 in decimal )

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Cam is 3 years older than Lara, If there combined age is 63, determine their ages by solving simultaneous elmination.

dem82 [27]

Answer:

30 and 33 years

Step-by-step explanation:

Let Lara be x years

Cam =x+3

x+x+3=63

2x+3=63

2x=63-3

2x=60

x=60/2

x=30

Lara=30years and Cam =30+3+33years

5 0

2 years ago

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

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2 years ago

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How to determine if a number is prime or composite

Nookie1986 [14]

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6 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

alex41 [277]

the correct answer should be (b) 6 is extraneous.

If you look at the second row of the solution you see (x-6) as a factor in one of the denominators. That's a problem: if 6 were a solution this denominator would become 0. But denominators being 0 is bad (undefined), so while 6 comes out of the algebraic manipulation as a solution, it is invalid, or extraneous.

-3 is a fine solution.

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2 years ago

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Look at the picture below .thx

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2 years ago