Answer:
The distance is given by |8-(-4)|=|8+4|=|12|=12.
Step-by-step explanation:
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Answer:
A and E
Step-by-step explanation:
Given
20x² - 26x + 8 = 0 ( divide through by 2 )
10x² - 13x + 4 = 0
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 10 × 4 = 40 and sum = - 13
The factors are - 5 and - 8
Use these factors to split the x- term
10x² - 5x - 8x + 4 = 0 ( factor the first/second and third/fourth terms )
5x(2x - 1) - 4(2x - 1) = 0 ← factor out (2x - 1) from each term
(2x - 1)(5x - 4) = 0
Equate each factor to zero and solve for x
2x - 1 = 0 ⇒ 2x = 1 ⇒ x = 
→ E
5x - 4 = 0 ⇒ 5x = 4 ⇒ x = 
→ A
Answer:
HJ = 17
IK = 30
Step-by-step explanation:
to find HI :
GH : 2
HI : ?
IJ : 12
GI : 7
HI = GI - GH
HI = 7- 2
HI = 5
HI + IJ = HJ
5 + 12 = HJ
17 = HJ
To find IK
IJ = ?
JK = 12
KL = ?
IL = 49
JL = 31
To find IK we need to find KL and IJ
KL = JL - JK
KL = 31 - 12
KL = 19
IJ = IL - (JK + KL)
IJ = 49 - 31
IJ = 18
IK = IJ + JK
IK = 18 + 12
IK = 30
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
