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dalvyx [7]
3 years ago
10

Hey i need help (winner gets crowned brainlyest)

Mathematics
1 answer:
Alenkinab [10]3 years ago
3 0

Answer:

Step-by-step explanation:

13) x⁴-12x² +36    

(a-b)² = a²-2ab+b²

a = x² ; b = 6

(x²)² - 2 * x² * 6 + 6² =  (x² - 6)²

14) w⁴- 14w² - 32 =     w⁴+ 2w² - 16w² - 32  = w² (w² + 2) - 16 (w²+2)

                           = (w² + 2) (w² -16 )

15) k³ + 7k² - 44k = k ( k² + 7k -44)  = k ( k+11 ) ( k-4 )

16) 2a³ +28a²+96a =2a(a²+14a+48) = 2a(a+6)(a+8)

17) -x³ +4x² +21x = (-x) ( x² - 4x - 21) = (-x)(x-7)(x+3)

18) m⁶ - 7m⁴ -18m² = m² ( m⁴-7m²-18) = m² (m²-9)(m²+9)

                             = m² (m+1) (m-1)(m²+9)

19) 9y⁶ +6y⁴ + y²= y² ( 9y⁴+6y²+1) = y² (3y²+1)²

20) 8c⁴+10c² -3 = (4c +1)(2c-3)

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Please Solve this, it would be extremely helpful for me.
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Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\

Cross multiply

EF * 15 = 27 * 7.5

EF =\frac{27*7.5}{15}\\\\

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}

AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm

AC = 37.5 cm

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2 years ago
The circumference of the smaller circle is 30% of the circumference of the larger circle. Find the circumference of the larger c
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The circumference of the larger circle is 94.2 inches.

Step-by-step explanation:

8 0
2 years ago
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What is the equation for a circle with a center at (-2,-4) that passes through the point (3,8)?
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Check the picture below.

so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
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\end{array}\qquad
%   coordinates of midpoint 
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\\\\\\
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\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}~~~ 
%  distance value
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\\\\\\
d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2}
\\\\\\
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\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{\frac{1}{2}}{ h},\stackrel{2}{ k})\qquad \qquad 
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\\\\\\
\left( x-\frac{1}{2} \right)^2+(y-2)^2=\left( \frac{13}{2} \right)^2\implies \left( x-\frac{1}{2} \right)^2+(y-2)^2=\frac{169}{4}

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