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3 years ago

In the image, point A marks the center of the circle. Which two lengths must form a ratio of 2:1?

Mathematics

2 answers:

Elza [17]3 years ago

6 0

Answer:

option b. $EF:AD=2:1$

Step-by-step explanation:

we know that

In a circle the ratio of the diameter to the radius is equal to $2:1$

because

$D=2r$

$2r/r=2:1$

In this problem

EF is a diameter

AD is a radius

therefore

$EF:AD=2:1$

harina [27]3 years ago

5 0

I hope this helps you

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The product of two even numbers is always divisible by 4 true or false

never [62]

Answer: TRUE

Step-by-step explanation:

an even number is divisible by 2.

Consider even number j which equals 2a and even number k which equals 2b. Then the product of these two numbers (j and k) = 2a*2b = 4ab.

4ab is divisible by 4.

BONUS:

The product of three even numbers is divisible by 2³

The product of four even numbers is divisible by 2⁴

The product of five even numbers is divisible by 2⁵

do you see the pattern?

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3 years ago

Which sets of numbers are closed under subtraction?

Alekssandra [29.7K]

I'm pretty sure it is only b because when you subtract two rational numbers, you always get back a rational number.

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3 years ago

In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg

siniylev [52]

Answer:

$\dfrac{AP}{PC}=\dfrac{3}{5}$

$\dfrac{BN}{CN}=\dfrac{3}{5}$

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

angle A is the common angle, so $\angle MAP\cong \angle DAC$ by reflexive property;
angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,$

$\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}$

Consider triangles ACB and PCN. In these triangles,

angle C is the common angle, so $\angle ACB\cong \angle PCN$ by reflexive property;
angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,$