All categories

2 years ago

(Redoing this question because it formatted wrong)

Mathematics

1 answer:

mojhsa [17]2 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

Put 3/8, 2/5, and 0.38 from least to greatest

Vesnalui [34]

3/8=0.375
2/5=0.4
0.38=0.38

0.375<0.38<0.4

therefore: 3/8,0.38,2/5

*Remember that you can convert fractions to decimals by dividing the numerator by the denominator! This makes it much easier to compare :)

5 0

2 years ago

Read 2 more answers

The number of customers that come to a certain clothing store each day follows a normal distribution. The mean number of custome

Lera25 [3.4K]

Answer:

c 2.5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Give one number that makes the inequality 3x < 25 true

elena55 [62]

1 is the safest answer.

4 0

3 years ago

Read 2 more answers

If 9x = 63, then x = ___. (Only input whole number) Numerical Answers Expected! Answer for Blank 1:

lina2011 [118]

If 9x = 63, then x = 7

5 0

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