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3 years ago

Given: AB ≅ BC and MA ≅ PC ∠AMO ≅ ∠CPO Prove: ∆AMO ≅ ∆CPO

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

5 0

Are AB BC MA PC in the triangle??

Schach [20]3 years ago

5 0

Answer:

OM = ON by using concurrency of Δs BOM and DON

Step-by-step explanation:

In the figure ABCD:

∵ AB // CD

∵ CB // AD

∵ In any quadrilateral If every two sides are parallel then

it will be a parallelogram

∴ ABCD is a parallelogram

∴ AC and BD bisects each other at O ⇒ (properties of parallelogram)

∴ OD = OB ⇒ (1)

∵ BC // AD

∴ m∠CBD = m∠ADB ⇒ alternate angles

∵ M ∈ BC , N ∈ AD , O ∈ BD

∴ m∠MBO = m∠NDO ⇒ (2)

∵ BD intersects MN at O

∴ m∠MOB = m∠NOD ⇒ (3) (vertically opposite angles)

From (1) , (2) and (3)

∴ ΔBOM ≅ ΔDON

∴ OM = ON

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Ricky has 3/4 of a bag of Skittles. He wants to split the rest between between his 5 friends. What fraction of the bag will each

Wittaler [7]

Answer:

3/20, with the assumptions below.

Explanation:

The question seems incomplete.

In order to show you the procedure, I make some assumptions.

Assuming that Ricky wants to split the full content of the 3/4 of a bag between his 5 friends, to find what fraction of the bag will each friend receive you must divide 3/4 by 5.

The operation is:

$\dfrac{3}{4}\div 5$

Convert the whole number 5 into fraction using 1 as denominator:

$\dfrac{3}{4}\div \dfrac{5}{1}$

Transform the division into multiplication changing the divisor into its reciprocal:

$\dfrac{3}{4}\times \dfrac{1}{5}$

Multiply numerator with numerator and denominator with denominator:

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7 0

3 years ago

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antiseptic1488 [7]

Answer:

64/49

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(bc)

(a^b)(a^c) = a^(b+c)

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a^0 = 1

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Using these rules, we can simplify the given expression to ...

$\left(\dfrac{8\cdot 4\cdot 2}{8\cdot 7}\right)^2\times\left(\dfrac{8^0}{7^{-3}}\right)^3\times 7^{-9}\\\\=\dfrac{8^2}{7^2}\times(8^{0\cdot 3}\cdot 7^{3\cdot 3})\times 7^{-9}\\\\=\dfrac{64}{49}\times(8^07^{9-9})=\boxed{\dfrac{64}{49}}$