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3 years ago

HELPPPPPP !!! I seriously don't know what to write.

Mathematics

1 answer:

tresset_1 [31]3 years ago

3 0

Answer:

This question was previously answered on Brainly...

Be sure to put this into your own words.

Answer: check explanation for the solution

Step-by-step explanation:

A business that offers services to people by providing many amusement and fun with gate fees

The algebraic equations to be used is the general linear equation

Y = MX + C

Where

Y = total income or money realised

M = rate or price rate

X = number of goods or services

C = flat rate or gate fees

The business can also operate differently by using exponential equation

A = P(1 + R%)^t

Where

A = profit

P = capital

R = rate

t = time

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F(x)=- x ^ 2 - 1,x ne5\\ -3,x=5 lim x -> 5 f(x) = lim x -> 5 f(x) Find if

Wewaii [24]

Answer:

the answer to the question is 1

3 0

2 years ago

Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by $\sqrt{x}$

$b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x} }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\$

Now, find a +b

$a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}$

Combine like terms

$= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}$

Now find (a + b)²

(a +b)² = $(\frac{2\sqrt{x} }{x})^{2}$

$= \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}$

Hint: $\sqrt{x} *\sqrt{x} =\sqrt{x*x}=x$

5 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .