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timama [110]
2 years ago
7

Eeeeeeeeeeeeeeeeeeeeeeeeee

Mathematics
2 answers:
Blizzard [7]2 years ago
8 0

Answer:

well said

Step-by-step explanation:

nice

Airida [17]2 years ago
7 0

Answer:

I agree

Step-by-step explanation:

Yes

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4. CSC X COS X - 2 cos x = 0
Alisiya [41]

Answer:

infinitely many answers, any odd integer n times π/2

ex. π/2, 3π/2, 5π/2

Step-by-step explanation:

we know that cos (π/2) or cos(3π/2) is 0

if one factor is 0, the product is zero, so we will not care about csc(x).

x is any odd integer n times π/2

3 0
3 years ago
-4x^2+20-4 in standard form
oksian1 [2.3K]
The answer should be -4x^2+16               
4 0
3 years ago
3x2 = 81<br> b. (x - 1)(x + 1)-9 = 5x<br> C. X - 9x + 10 = 32<br> d. 6x(x - 9) = 29
irakobra [83]

Answer:

a. 3x^2 = 81

  x^2 = 27

x =  3\sqrt{3} or -3\sqrt{3}

Step-by-step explanation:

6 0
3 years ago
The graph shows the solution to which system of equations?
vichka [17]
Start by seeing where the lines go. They cross (intersect) at (1, -1). We can use this to check later.
Now for slope: green is -4/2 = -2
pink is 4/2 = 2
Now we can create the equations
Let's make green = g(x) and pink = p(x)
if we use y = mx + b, then the green has a y-intercept (b) of +1
So g(x) = -2x + 1
pink has a y-intercept of -3, so p(x) = 2x - 3
Now let's plug n play: put our solutions x into each equation and confirm that it makes the y = -1
g(x) = -2x + 1 = -2(1) + 1 = -2+1 = -1
p(x) = 2x - 3 = 2(1) - 3 = 2-3 = -1

✔ YES THEY ARE CONFIRMED
6 0
3 years ago
Read 2 more answers
Write an equation of the line that passes through (18, 2) and is parallel to the line 3y−x=−12
miskamm [114]

keeping in mind that parallel lines have the same exact slope, hmmmm what's the slope of the line above anyway?

\bf 3y-x=-12\implies 3y=x-12\implies y=\cfrac{x-12}{3}\implies y = \cfrac{x}{3}-\cfrac{12}{3} \\\\\\ y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{3}}x-4\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

so we're really looking for the equation of a line whose slope is 1/3 and runs through (18,2)

\bf (\stackrel{x_1}{18}~,~\stackrel{y_1}{2})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{18}) \\\\\\ y-2=\cfrac{1}{3}x-6\implies y=\cfrac{1}{3}x-4

3 0
3 years ago
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