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katrin2010 [14]
2 years ago
7

Answer these problems please! Solving for x!

Mathematics
1 answer:
Tema [17]2 years ago
3 0

Answer:

4. x= 26 5.  7.333333 6. 16.4

Step-by-step explanation:

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To find the x intercept set y = 0

x^2 + x  - 2 =0 

(x+2)(x-1) = 0

x = -2 , x = 1


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Last question for today ✌
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I think it is D

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QUESTION<br> Learning Page<br> AABC and APQR are similar. Find the missing side length.
melisa1 [442]

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3 0
3 years ago
Find the solution of the following equation whose argument is strictly between 270^\circ270 ∘ 270, degree and 360^\circ360 ∘ 360
Natasha2012 [34]

\rightarrow z^4=-625\\\\\rightarrow z=(-625+0i)^{\frac{1}{4}}\\\\\rightarrow x+iy=(-625+0i)^{\frac{1}{4}}\\\\ x=r \cos A\\\\y=r \sin A\\\\r \cos A=-625\\\\ r \sin A=0\\\\x^2+y^2=625^{2}\\\\r^2=625^{2}\\\\|r|=625\\\\ \tan A=\frac{0}{-625}\\\\ \tan A=0\\\\ A=\pi\\\\\rightarrow z= [625(\cos (2k \pi+pi) +i \sin (2k\pi+ \pi)]^{\frac{1}{4}}\\\\k=0,1,2,3,4,....\\\\\rightarrow z=(625)^{\frac{1}{4}}[\cos \frac{(2k \pi+pi)}{4} +i \sin \frac{(2k\pi+ \pi)}{4}]

\rightarrow z_{0}=(625)^{\frac{1}{4}}[\cos \frac{pi}{4} +i \sin \frac{\pi)}{4}]\\\\\rightarrow z_{1}=(625)^{\frac{1}{4}}[\cos \frac{3\pi}{4} +i \sin \frac{3\pi}{4}]\\\\ \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]\\\\ \rightarrow z_{3}=(625)^{\frac{1}{4}}[\cos \frac{7\pi}{4} +i \sin \frac{7\pi}{4}]

Argument of Complex number

Z=x+iy , is given by

If, x>0, y>0, Angle lies in first Quadrant.

If, x<0, y>0, Angle lies in Second Quadrant.

If, x<0, y<0, Angle lies in third Quadrant.

If, x>0, y<0, Angle lies in fourth Quadrant.

We have to find those roots among four roots whose argument is between 270° and 360°.So, that root is

   \rightarrow z_{2}=(625)^{\frac{1}{4}}[\cos \frac{5\pi}{4} +i \sin \frac{5\pi}{4}]

5 0
3 years ago
Please help me guys
Dima020 [189]
Hello!

OC = OD = Radius
6x + 2 = 10x - 16
6x - 10x = -16 - 2
-4x = -18
x = -18/-4
x = 4.5

Radius = OC = 6(4.5) + 2 = 29 units
5 0
3 years ago
Read 2 more answers
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