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Alexxandr [17]
3 years ago
11

Circle the ordered pairs that are solutions to the inequality y - 2x < - 1.

Mathematics
1 answer:
TEA [102]3 years ago
5 0
All we have to do, is plug in the ordered pairs, and see which ones are true to the equation. (x,y)

(1,0) After plugging those into the equation we get: 2<-1. this is not true, so this answer is false.

I continued this all the way down the line:

(0,1)= 1<-1 = false
(0,-1)=-1<-1=false
(-1,0)= -2<-1=true
(3,2)=-1<-1=false
(-2,-4)=2<-1=false
(2,4)=2<-1=false
(-4,2)=8<-1=false
(0,0)=0<-1=false

There is only one true ordered pair for this equation:

(-1,0)


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In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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3 years ago
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pantera1 [17]
Log(3^2)+log(5)=log(x)
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Given: Sine (A) = four-fifths, StartFraction pi Over 2 EndFraction &lt; A &lt; Pi and Sine (B) = StartFraction negative 2 StartR
ELEN [110]

Answer:

D

Step-by-step explanation:

Did test(edge 2020)

8 0
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inn [45]

Answer:

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8. Solve x/10 = –7. A. x = –0.7 B. x= –17 C. x = –70 D. x = 3
ANEK [815]
For this problem, I believe you would just have to plug in for x.

For example, -17 divided by 10 and so on.

The answer should be C, -70. -70 divided by 10 equals -7. 
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