Because of the transversal and the parallel lines, we know that A+B=180°.
So:
5x + 20 + 9x - 92 = 180
14x - 72 = 180
14x = 180 + 72
14x = 252
x = 252/14
x = 18
A = 5x + 20
A = 5*18 + 20
A = 90 + 20 = 110
(Let’s check:
B = 9*18 - 92
B = 162 - 92
B = 70
70 + 110 = 180)
Answer:
complex number
Step-by-step explanation:
77i is an imaginary number but once you put that 3 in the equation it turns into a complex number bc its more than just the imaginary now
For every 1,000 feet you go up in elevation, the temperature decreases by about 3.3°F
Given:
Temperature at the bottom of the mountain is 74°F
The top of the mountain is 5500 feet above you
Change (reduction) in temperature is 3.3 X 5500 / 1000
= 18.15°F
Temperature at the top of the mountain is 74°F – 18.15°F
= 55.85°F
Answer:
1st problem:
-1, 2i, -2i
Set the function equal to 0 and divide for x
0=x^3+x^2+4x+4
Subtract 4 from both sides leaving you with
-4=x^3+x^2+4x
divide both sides by four
-4÷4= -1
That is how you get -1
you now have -1=x^2+x^3+x
Solve for x buy finding the square roots of x^2, x^3 and subtracting x from both sides and you get 2i, -2i
Your answer to number one is -1, 2i, -2i
Second problem:
Find the LCD to be able to add these first
The LCD of (x-4)/x^2-2x is
x(x-2)
That is the first side done
The second side's LCD is (x+2)(x-2)
Now you can add the two together
x(x-2)+(x+2)(x-2)=
Your answer in fraction form is
(x+4)/x(x+2)
A rhombus has four congruent sides. Without loss of generality, we denote the rhombus by the consecutive points PQRS.
1. Draw diagonal PR.
2. Consider triangle PQR. We conclude that PQ=RQ because a rhombus has four congruent sides. Triangle PRQ is isosceles => angles QPR and QRP are congruent.
3. Similarly, PS=RS. Triangle PSR is isosceles.
4. PR (diagonal) is common side, so triangles PQR is congruent to PSR (SSS).
5. Therefore angles QRP=QPR=SPR=SRP.
6. Since angles QPR=SPR, and SPR=QPR, we conclude that the diagonal PR is an angle bisector.
Finally, using similar arguments, we can show that diagonal QS bisects angles at Q and S.