All categories

3 years ago

Given circle O with chords CD = 12 and

Mathematics

1 answer:

LuckyWell [14K]3 years ago

0 0

Answer:

FB=6 Units
OE= 8 Units

Step-by-step explanation:

Theorem: A line drawn from the center perpendicular to the chord bisects the chord.

By the Theorem above, Point E and F bisects CD and AB into 6 Units each as seen in the diagram.

Therefore:

|FB|=6 Units

In Right Triangle OED, applying Pythagoras Theorem

$10^2=6^2+|OE|^2\\|OE|^2=100-36\\|OE|^2=64\\|OE|^2=8^2\\|OE|=8 \:Units$

Therefore:

FB=6 Units
OE= 8 Units

You might be interested in

Help!! Will give brainliest

patriot [66]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Identify the glide reflection rule in the given figure

Maslowich

Answer:

Option (3)

Step-by-step explanation:

Glide reflection of a figure is defined by the translation and reflection across a line.

To understand the glide rule in the figure attached we will take a point A.

Coordinates of the points A and A' are (2, -1) and (-2, 4).

Translation of pint A by 5 units upwards,

Rule to be followed,

A(x, y) → A"[x, (y + 5)]

A(2, -1) → A"(2, 4)

Followed by the reflection across y-axis,

Rule to be followed,

A"(x, y) → A'(-x, y)

A"(2, 4) → A'(-2, 4)

Therefore, by combining these rules in this glide reflections of point A we get the coordinates of the point point A'.

Option (3) will be the answer.

4 0

3 years ago

Factor 1/3 out of 1/3x−50/3

wlad13 [49]

Answer:

50 1/3

Step-by-step explanation:

3 0

3 years ago

Solve for x? Please

sasho [114]

Check the picture below.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al

Furkat [3]

Answer: $\frac{\sqrt[4]{10xy^3}}{2y}$

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

$\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\$

The idea is to get something of the form $a^4$ in the denominator. In this case, $a = 2y$

To be able to reach the $16y^4$ , your teacher gave the hint to multiply top and bottom by $2y^3$

For more examples, search out "rationalizing the denominator".

Keep in mind that $\sqrt[4]{(2y)^4} = 2y$ only works if y isn't negative.

If y could be negative, then we'd have to say $\sqrt[4]{(2y)^4} = |2y|$ . The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0

2 years ago