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2 years ago

Given that V=1,3,4,2 and W= 6,4,7,2 then VW is

Mathematics

1 answer:

Rus_ich [418]2 years ago

4 0

Answer:

2,4 possibly

Step-by-step explanation:

if its V (intersection) W

then its 2,4

if it has something to do with patterns then i dont know

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MAVERICK [17]

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3 years ago

(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this

Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta$

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

$\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta$

Simplify. The denominator uses the difference of two squares pattern:

$\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta$

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

$\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta$

Split into two separate fractions:

$\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta$

Rewrite the two fractions:

$\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta$

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

$\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta$

Hence verified.

8 0

2 years ago

1. Simplify: 50r2-20r - 10r A. -5r + 2 B. -5r-2 C. 5r2 +2

mel-nik [20]

Answer:50r2−30r

Step-by-step explanation:

got it of the web.

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2 years ago

Find the x and y intercepts for the equation 3/5x+1/2y=30

Jobisdone [24]

3/5x + 1/2y = 30
-3/5x. -3/5x
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—————————
1/2
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2 years ago

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NemiM [27]

Answer: y = (x - 5)² - 1

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3 years ago