What is the place of 2 in 24,398,407,268

Mathematics

1 answer:

kirill115 [55]3 years ago

8 0

Well they are two number twos there the first on would be ten billion and the second would be hundreds.

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Trigonometric identities assignment 4

kykrilka [37]

Answer:

Answer: choice D.

Step-by-step explanation:

Trigonometric Functions

We need to recall some properties and definitions of the trigonometric functions.

$\displaystyle \tan \theta=\frac{\sin\theta}{\cos\theta}$

$\displaystyle \sec \theta=\frac{1}{\cos\theta}$

We need to simplify the following expression:

$5\sin\theta\sec\theta$

Substituting the identity for the secant:

$\displaystyle 5\sin\theta\sec\theta=5\sin\theta.\frac{1}{\cos\theta}$

Operating:

$\displaystyle 5\sin\theta\sec\theta=5\frac{\sin\theta}{\cos\theta}$

Substituting the definition of the tangent:

$\displaystyle 5\sin\theta\sec\theta=5\tan\theta$

Correct choice D.

6 0

3 years ago

HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x

Anna35 [415]

$\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0$

$\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}$