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3 years ago

7w + (-11w) help i suck at math thank you

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

6 0

Answer:

-4

Step-by-step explanation:

7w + (-11w)

Postive negative negative

7w-11w

= - 4

Genrish500 [490]3 years ago

3 0

=4 the answer is your welcome

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beks73 [17]

The result of expanding the trigonometry expression $\sin^2(\theta) * (1 + \cos(\theta))$ is $cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

<h3>How to evaluate the expression?</h3>

The expression is given as:

$\sin^2(\theta) * (1 + \cos(\theta))$

Express $\sin^2(\theta)$ as $1 - \cos^2(\theta)$ .

So, we have:

$\sin^2(\theta) * (1 + \cos(\theta)) = (1- \cos^2(\theta)) * (1 + \cos(\theta))$

Open the bracket

$\sin^2(\theta) * (1 + \cos(\theta)) = 1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Express 1 as cos°(Ф)

$\sin^2(\theta) * (1 + \cos(\theta)) = cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Hence, the result of expanding the trigonometry expression $\sin^2(\theta) * (1 + \cos(\theta))$ is $cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

Read more about trigonometry expressions at:

brainly.com/question/8120556

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3 0

2 years ago

Which of the following is the inverse of y = 12 Superscript x? y = log Subscript one-twelfth Baseline x y = log Subscript 12 Bas

wariber [46]

Answer:

y = lnx/ln12

Step-by-step explanation:

Given the function y = 12^x, to find the inverse of the function, we news to write x as a function of y as shown:

Given y = 12^x

Taking ln of both sides

ln y = ln 12^x

ln y = xln12

Divide both sides by ln12

lny/ln12 = xln12/ln12

x = lny/ln12

x = ln(y-12)

Replacing y with x and x with y

The inverse of the function will be:

y = lnx/ln12

3 0

3 years ago

Solve the differential equation. (y^2 + xy^2)y' = 1

stira [4]

$(y^2 + xy^2)y' = 1 \ \Rightarrow\ (y^2 + xy^2)\frac{dy}{dx}= 1 \ \Rightarrow\\ \\
y^2(1 + x) \frac{dy}{dx} = 1 \ \Rightarrow\ y^2 dx = \frac{dx}{1 + x}\ \Rightarrow \\ \\
\int y^2 dx =\int \frac{dx}{1 + x} \ \Rightarrow\ \frac{1}{3}y^3 = \ln|1+x| + C\ \Rightarrow \\ \\
y^3 = 3 \ln|1 + x| + 3C\ \Rightarrow\ y = \sqrt[3]{3\ln|1+x| + K},\ \text{where $K = 3C$}
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