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2 years ago

Which of the following segments is a radius of 0?

Mathematics

2 answers:

erastovalidia [21]2 years ago

8 0

Not sure i remember this but i think radius is like half of the diameter. it’s a line from the center, to a side of the circle. in this situation, line RO would be representing the radius.

answer: D, line RO
hope this helps :)

Brut [27]2 years ago

5 0

Answer:

Step-by-step explanation:

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Consider a situation in which P(A) = 1/8, P(C) = 1/4, and P(A and B) = 1/12. What is P(B and C)?

V125BC [204]

ANSWER

$P(B \: and \: C) = \frac{1}{6}$

EXPLANATION

Recall that,

$P(A \: and \: B)=P(A) \times P(B)$

But we were given that,

$P(A)= \frac{1}{8}$

and

$P(A \: and \: B) = \frac{1}{12}$

We substitute these values into the above formula to obtain,

$\frac{1}{12} = \frac{1}{8} \times P(B)$

This implies that,

$\frac{1}{12} \times 8=8 \times \frac{1}{8} \times P(B)$

This simplifies to,

$P(B) = \frac{8}{12}$

$P(B) = \frac{2}{3}$

So we can now find
$P(B \: and \: C).$

We use the same formula again,

$P(B \: and \: C) = P(B) \times P(C)$
We substitute the values to get,

$P(B \: and \: C) = \frac{2}{3} \times \frac{1}{4}$
We multiply out to get,

$P(B \: and \: C) = \frac{1}{6}$

7 0

3 years ago

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What are the coordinates of point P on the coordinate grid below?

Leona [35]

The coordinates are (-3/8, 5/8), work is shown in the picture attached

5 0

2 years ago

Please help me with this , probably easy for others but i don't got energy to do it : ')

Simora [160]

Answer:

I believe its B

Step-by-step explanation:

7 0

2 years ago

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If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

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1 year ago

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PLEASE HELP ASAP! Will give BRAINLIEST! Please answer correctly!<br><br> No guessing!

Klio2033 [76]

Answer:

c.)

Step-by-step explanation:

i graphed it on my calculator