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2 years ago

Here xchx solve this mate :)Thanks. 12 points smh

Mathematics

1 answer:

NikAS [45]2 years ago

4 0

Answer:

1. y = (-4^3)^2

2. y= 2^4/ 6^2

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(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

Please answer correctly !!!!!!! Will mark brainliest answer !!!!!!!!

Allisa [31]

Answer:

A is the answer i believe

Step-by-step explanation:

3 0

3 years ago

Does anyone know how to get the quadratic formula on a ti-84 calculator or any formulas for the ACT?

adoni [48]

Answer:u 2 look sooooooooooo hot ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

Step-by-step explanation:

4 0

3 years ago

Solve please.....Thank you

Harlamova29_29 [7]

Answer:

● $\displaystyle 1$

Step-by-step explanation:

According to the Definition of Rational Exponents [part I], you can rewrite radicals as exponents:

$\displaystyle a^{\frac{m}{n}} = \sqrt[n]{a}^m \\ \\ \sqrt{y^2} = y, where\:any\:invisible\:exponent\:is\:ALWAYS\:1$

I am joyous to assist you anytime

5 0

3 years ago

BRAINLIST AND A THANK YOU AND 5 stars WILL BE REWARDED PLS ANSER

vichka [17]

Answer:

The first picture's answer would be (6, 21)

Step-by-step explanation:

You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.

3 0

3 years ago