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OlgaM077 [116]
3 years ago
11

Determine the number of real solutions of -2x^2+ 5x – 3 = 0. 2 1 (double root) 5 0

Mathematics
1 answer:
Svetllana [295]3 years ago
5 0

Answer:

{ \tt{ - 2 {x}^{2} + 5x - 3 = 0 }} \\   { \tt{2 {x}^{2}  - 5x + 3 = 0}} \\ { \tt{(2x - 3)(x - 1) = 0}} \\ { \tt{x =  \frac{3}{2}  \: and \: 1}}

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Given that , the _____ justifies the conclusion that ∆ABD ≅ ∆AEC.
trasher [3.6K]
AAS Postulate

It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.

Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate. 

Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!

3 0
3 years ago
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PLEASE HELP ASAP. IM TIMED​
ICE Princess25 [194]

It will be a parallelogram.

8 0
3 years ago
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If P is the incenter of triangle JKL, find the measure of JKP.
ladessa [460]

Answer:

m∠JKP = 31.5°

Step-by-step explanation:

Incenter of a triangle is the point where all the bisectors of interior angles intersect each other.

JN is the angle bisector of ∠KJL.

Therefore, m∠KJN = m∠LJN

(7x - 6) = (5x + 4)

7x - 5x = 6 + 4

2x = 10

x = 5

m∠KJN = (7x - 6)

            = 7(5) - 6

            = 35 - 6

            = 29°

In ΔKJN,

m∠JKN + m∠KNJ + m∠NJK = 180°

m∠JKN + 90° + 29° = 180°

m∠JKN = 180°- 119° = 61°

Since KO is the angle bisector of ∠JKN,

m∠JKP = \frac{1}{2}(\angle JKN)

            = \frac{1}{2}(61)

            = 30.5°

6 0
3 years ago
I need help with this also I had was to repost it again because someone deleted it
igomit [66]

Answer: 8:05

Step-by-step explanation:

8 0
2 years ago
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Mr suresh had 7/9 kg of sugar. He used 1/3 of it. How much sugar did he use?​
BabaBlast [244]

1/3 * 7/9 kg = (1 * 7)/(3 * 9) kg = 7/27 kg

He used 7/27 kg of sugar.

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3 years ago
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