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n200080 [17]
3 years ago
11

The equation ( ) has no solution.

Mathematics
2 answers:
Inessa [10]3 years ago
8 0

Answer:

5(2.2y + 3.4) = 5(y - 2) + 6y

Step-by-step explanation:

There are no values of  y  that make the equation true.

No solution

So it is this one!

<u></u>

<u>Hope this helps! </u>

<u>Please Mark Brainliest!</u>

trasher [3.6K]3 years ago
6 0

Answer:

i dont see the equation

Step-by-step explanation:

sorry

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In Lisa's class 24 of the students are tall and 12 are short. In Paul's class 15 students are tall and 20 are short. Which class
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Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM
denis23 [38]

Answer:

The length of side KM is \sqrt{109} units.

Step-by-step explanation:

Given information:  KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3\sqrt{5}.

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is 3\sqrt{5}.

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

\frac{3x+5x}{2}=8

\frac{8x}{2}=8

x=2

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

KN=KA+AE+EN

10=6+2(EN)                (KA=EN, isosceles trapezoid)

EN=2

KE=KN-EN=10-2=8

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

Hypotenuse^2=base^2+perpendicular^2

(KM)^2=(KE)^2+(ME)^2

(KM)^2=(8)^2+(3\sqrt{5})^2

KM^2=64+9(5)

KM=\sqrt{109}

Therefore the length of side KM is \sqrt{109} units.

8 0
3 years ago
Read 2 more answers
Factor completely 2x2 + 4x − 2.
Fiesta28 [93]
Is that 2x or 2*2?
i need to know so i can factor it

8 0
4 years ago
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