Ask question

Ask question

All categories

3 years ago

11

Solve the volume formula V=lwh for h.

1 answer:

Alexeev081 [22]3 years ago

4 0

V=lwh

v/lw=h

thats what it is.

You might be interested in

I need the answer to this

garri49 [273]

For both of the contests, the variable is the number of competitors which we can set as x.
Consequently, the equation should be 5.50x+34.60=4.25x+64.60

5.50x=4.25x+30

1.25x=30

x=24

The fee of the contests is the same when there are 24 competitors

5 0

3 years ago

James designed a game of chance which involves identical balls numbered from 1 to 50. He selected 5 balls at random and found th

Bad White [126]

It ......... is .......... 38 .........

3 0

2 years ago

What is 125% of 144?

Monica [59]

Answer:

180

hope this helps :)

3 0

2 years ago

Read 2 more answers

What is the next number in the pattern?

forsale [732]

Answer:

9

Step-by-step explanation:

3 3/4= 3.75.

3.75-2=1.75

1.75+5.5=7.25 or 7 1/4

7.25+1.75=9

5 0

2 years ago

Read 2 more answers

(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