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3 years ago

In quadrilateral ABCD, AD ∥ BC. What must the length of segment AD be for the quadrilateral to be a parallelogram?

Mathematics

1 answer:

FrozenT [24]3 years ago

6 0

Answer: The Answer is C) 31 Units

Step-by-step explanation:

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20 feet. The builder asks him to give the measurements in yards. What will be the length in yards?Immersive Reader

Sedaia [141]

Answer:

6.67 yards

Step-by-step explanation:

20 feet. The builder asks him to give the measurements in yards. What will be the length in yards?

Please note that:

1 foot = 0.333 yard

20 feet = x yards

Cross Multiply

1 foot × x yards = 20 feet × 0.333 yards

x yards = 20 feet × 0.333 yards/1 foot

x yards = 6.66667 yards

Approximately = 6.67 yards

Therefore, the measurements in yards is 6.67 yards.

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2 years ago

Is the formula x+1=3/4y linear?

pentagon [3]

Yes, it is.
Formula X+1=3/4y is linear.

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2 years ago

PLEASE HELPPP!!!<br><br><br> THIS SHOULD BE EASY!

cupoosta [38]

Answer:

yes she can

Step-by-step explanation:

6 0

2 years ago

Mind giving me a hand?

zaharov [31]

Answer:

Jimmy has one candy times, 35 candies that his mom just gave him. He said, he's going togive 14 out of the 16 candies, he put a side to his mom

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2 years ago

Solve for x. Round your answer to the nearest hundredth.

Zarrin [17]

Answer:

Option B is correct

Step-by-step explanation:

2x² - 6x + 3 = 0

Using Quadratic formula:

x = $\frac{-b+-\sqrt{b^{2}-4ac}}{2a}$

a = 2, b = -6, c = 3

x = $\frac-(-6)+-{\sqrt{(-6)^{2}-4(2)(3)}}{2(2)}$

x = $\frac{-(-6)+-\sqrt{(-6)^{2}-4(2)(3)}}{2(2)}\\$

$=\frac{6+-\sqrt{36-24}}{4}\\\\=\frac{6+-\sqrt{12}}{4}\\\\=\frac{6+-3.46}{4}\\\\$

$=\frac{6+3.46}{4}$

x = 2.37

$=\frac{6-3.46}{4}$

x = 0.63

5 0

2 years ago

Read 2 more answers