All categories

3 years ago

TRUE or FALSE: In the quarter circle, the radius is 4 cm and the diameter is 8 cm.

Mathematics

1 answer:

viktelen [127]3 years ago

6 0

Answer: False.

Step-by-step explanation:

There does not exist a "quarter circle" as a circle with a radius of 4 units, the only notable circle that does exist is the unit circle, that is the circle where the radius is equal to 1, represented by the equatin x^2 + y^2 = 1

The term "quarter circle" actually does refer to a fourth part of a circle, not to a circle of radius 4.

So the statement is false

You might be interested in

Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo

mart [117]

Answer:

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

Step-by-step explanation:

Given:

$y = -6x^2 + 3x + 2$

Required

Rewrite in vertex form

The vertex form of an equation is in form of: $y = a(x - h)^2+ k$

Solving: $y = -6x^2 + 3x + 2$

Subtract 2 from both sides

$y - 2 = -6x^2 + 3x + 2 - 2$

$y - 2 = -6x^2 + 3x$

Factorize expression on the right hand side by dividing through by the coefficient of x²

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

$y - 2 = -6(x^2 - \frac{3x}{6})$

$y - 2 = -6(x^2 - \frac{x}{2})$

Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

<em>Rough work</em>

The coefficient of x is $\frac{-1}{2}$

It's square is $(\frac{-1}{2})^2 = \frac{1}{4}$

Adding inside the bracket of $-6(x^2 - \frac{x}{2})$ to give: $-6(x^2 - \frac{x}{2} + \frac{1}{4})$

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: $-6(\frac{1}{4})$

------------------------------------------------------------------------------------

The expression becomes

$y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

$y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})$

Factorize the expression on the right hand side

$y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2$

$y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2$

$y - (\frac{7}{2})= -6(x - \frac{1}{2})^2$

$y +\frac{7}{2} = -6(x - \frac{1}{2})^2$

Make y the subject of formula

$y = -6(x - \frac{1}{2})^2 -\frac{7}{2}$

<em>Solved</em>

7 0

3 years ago

Read 2 more answers

SOMEONE PLEASE HELP ME! I promise i will mark brainlest, but i need the right answer for this. help is much appreciated <3

tester [92]

Answer:

the third choice

Step-by-step explanation:

I haven't done those questions in a long time so I would double check. But I think its choice C because 0 has two functions, and I'm pretty sure that a number can only have 1 output, therefore C is not a function..

5 0

2 years ago

You roll a six-sided die twice. What is the probability of rolling a number divisible by 2 and then a number divisible by 3?

Aneli [31]

If I am rolling a single dice what is the probability of rolling a number that is divisible by 2 and 3? The only number divisible by 2 and 3 is 6. So, would the answer be 1 out 6 or 1/6

3 0

3 years ago

Read 2 more answers

Students in a large district's two high schools are offered a "Second Breakfast' program after their first-period class.

murzikaleks [220]

Answer:

Step-by-step explanation:

3 0

3 years ago

HII i really need help

Kipish [7]

Answer: