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3 years ago

Find the reference angle for a rotation of 202°

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

3 0

Answer:

I think it is 158°

Step-by-step explanation:

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This table gives a few (x,y) pairs of a line in the coordinate plane what is the y-intercept of the line (16,44) (24,64) (32,84)

lapo4ka [179]

Answer:

Step-by-step explanation:

find the equation of the line:

slope = rise / run

rise is 20, run is 8

20/8 = 5/2

slope intercept form is y = mx + b

where m is slope and b is y int

y = 5/2(x) + b

use a point given to find b

44 = 5/2(16) + b

44 = 40 + b

b = 4

5 0

3 years ago

Read 2 more answers

Please help show work I beg A 50 B 13 C 12 D 51

Viefleur [7K]

B. this answer would be B.

3 0

3 years ago

Let a and b be roots of x² - 4x + 2 = 0. find the value of a/b² +b/a²

erastovalidia [21]

Answer:

$\dfrac{a}{b^2}+\dfrac{b}{a^2}=10$

Step-by-step explanation:

Given equation: $x^2-4x+2=0$

The roots of the given quadratic equation are the values of x when $y=0$ .

To find the roots, use the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=1, \quad b=-4, \quad c=2$

$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$