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3 years ago

What is the value of x? Enter your answer in the box.

Mathematics

1 answer:

vaieri [72.5K]3 years ago

4 0

I think it’s 22, not sure tho

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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 root5 Find: KM

Alla [95]

Answer:

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

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, $\frac{LM}{KN}=\frac{3}{5}$ , 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 $\frac{LM}{KN}=\frac{3}{5}$ , 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 KM is $\sqrt{109}$ units.

5 0

3 years ago

You have to answer to other question before answering this one, just in case anyone was confused

Nuetrik [128]

Answer: C

Explanation: You first convert the mixed numbers into improper fractions ( including the ones inside and outside of the parentheses) Then, subtract the fractions inside of the parentheses. Then whatever you got, add it to the fraction that was outside of the parentheses.

(7^5/6 - 5^1/3) + 1^1/9
^
(47/6 - 16/3) + 10/9
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5/2 + 10/9

65/18 = 3^11/18

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3 years ago

Blank thousands equals 5,000 ones?

KIM [24]

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