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3 years ago

Find area triangle of 15 ft and 18 ft The area of the triangle is. Ft2

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

a= 135

Step-by-step explanation:

18*15=135

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Answer:

1,3,4

Step-by-step explanation:

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12. Which three correlation coefficents have a strong linear correlation.

timofeeve [1]

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3 years ago

Willow poured the same amount of water into 8 flower pots. She used 5 gallons of water. How much water did each flower get?

larisa86 [58]

Answer:

0.625 gallons each

Step-by-step explanation:

Because she used 5 gallons of water, and poured the same amount of water for each plant, divide the number of gallons she used by the number of plants:

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3 years ago

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G(x)= x squared -6/ 3x+10<br> g(4)= ?

andrezito [222]

Answer:

g(4) = 18

Step-by-step explanation:

You have to input 4 in the x poistion of the equation :

g(x) = x² - 6/3x + 10

Solve:

g(4) = 4² - 6/3(4) + 10

=> Multiply and distrubute

g(4) = 16 - 8 + 10

=> Smplify

g(4) = 18

-Chetan K

3 0

2 years ago

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=3root5 Find: KM

Sergio039 [100]

Answer:

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

Step-by-step explanation:

Given the statement: 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.

6 0

3 years ago