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Mathematics

1 answer:

Anvisha [2.4K]3 years ago

3 0

Step-by-step explanation:

$\displaystyle y = Asec(Bx - C) + D$

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

$\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{4} = 0 \\ Period → \frac{2}{4}π = \frac{π}{2}$

Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|$

NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the midline where the amplitude is. Moreover, ALL tangent, secant, cosecant, and cotangent functions have NO AMPLITUDE.

I am joyous to assist you anytime.

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What is the lateral area of the cone?

Agata [3.3K]

Your answer would be ~ 305.83 !!

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3 years ago

What is -√169? (note: negative in front of the √)

Mademuasel [1]

Hello!
Ignore the negative and solve the square root which is 13.
Add the negative with 13 you get -13
Special case is when it is square root of -169 then it will involved with i.
But for this question just solve it directly.
Have a great day!

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3 years ago

A polynomial expression is shown below.

tester [92]

The polynomial is (mx^3+3)(2x²+5x+2)-(8x^5 +20x^4)
if it is reduced to 8x^3+6x²+15x+6, so we can find the value of m

(mx^3+3)(2x²+5x+2)-(8x^5+20x^4) = 8x^3+6x²+15x+6
2mx^5+5mx^4+2mx^3+6x²+15x+6-8x^5-20x^4=8x^3+6x²+15x+6
2mx^5+5mx^4+2mx^3=8x^3+6x²+15x+6-6x²-15x-6+ 8x^5+20x^4
= 8x^5+20x^4+8x^3= 4(2x^5+5x^4+2x^3)
finally

m(2x^5+5x^4+2x^3)=4(2x^5+5x^4+2x^3), and after simplification

C: m=4

4. When the expression is factored x²-3x-18 completely,

one of its factor is x-6
x²-3x-18=0
D= 9-4(-18)= 81, sqrtD=9 x=3-9/2= -6/2= -3, and x=3+9 / 2= 6
so x²-3x-18= (x-6)(x+6)

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2 years ago

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vlabodo [156]

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3 years ago

Read 2 more answers

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0

3 years ago