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JulsSmile [24]
3 years ago
14

Helpppppppppppppooo

Mathematics
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer:

4/7

Step-by-step explanation:

35r-21<-35r+19

70r-21<19

70r<40

40/70

4/7

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Give statements and reasons
PolarNik [594]

Answer:

A two column proof is presented as follows;

Statement {}                                     Reason

1. AB ║ DE, BD bisects AE        {}    Given

2. ∠BAE = ∠AED, ∠ABD = ∠BDA {} Alternate ∠s are equal

3.  \overline {AC} =  \overline {EC} ,  \overline {BC} =  \overline {CD} {} {}            By definition of bisection of line AE by BD

4. ΔABC ≅ ΔEDC  {}                         By SAA, rule of congruency

Step-by-step explanation:

Step 1. AB ║ DE, BD bisects AE        {}    Given

Step 2. ∠BAE , ∠AED, and ∠ABD, ∠BDA {} are pairs of alternate angles formed by the parallel lines, AB and DE and are therefore, equal

Step 3.  \overline {AC} =  \overline {EC} ,  \overline {BC} =  \overline {CD}  The bisection of line gives two lines of equal length. The bisection of AE by BD gives,  \overline {AC} and  \overline {EC} where \overline {AC} =  \overline {EC}

Similarly, the bisection of BD by AE gives, \overline {BC} and  \overline {CD}, where  \overline {BC} =  \overline {CD}

Step 4. ΔABC ≅ ΔEDC  {}  By the Side-Angle-Angle (SAA), rule of congruency, which states that two triangles having two angles and the corresponding non included sides of each triangle equal to the other, the two triangles are congruent.

6 0
3 years ago
X-4(3-2x) =3(3x+4) is this many soultions no solutions
taurus [48]
No solutions. It would be: 9x-12=9x+12. The difference is that one of the 12’s is negative and one is positive. Hope this helps! Have a great day :)
4 0
3 years ago
Which diagram represents the hypothesis of the converse of corresponding angles theorem?
BabaBlast [244]

Answer:

the first diagram

Step-by-step explanation:

first one

5 0
3 years ago
Find the derivative of f(x) = 6x + 2 at x = 1.
earnstyle [38]
f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=6x+2;\ x_0=1\\\\f'(1)=\lim\limits_{x\to x_0}\dfrac{(6x+2)-(6\cdot1+2)}{x-1}=\lim\limits_{x\to x_0}\dfrac{6x+2-8}{x-1}\\\\=\lim\limits_{x\to x_0}\dfrac{6x-6}{x-1}=\lim\limits_{x\to x_0}\dfrac{6(x-1)}{x-1}=\lim\limits_{x\to x_0}6=6\\\\\\Answer:\boxed{f'(1)=6}
6 0
3 years ago
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