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PSYCHO15rus [73]

2 years ago

10

Circle G and X connect at point Y. The length of G Y is 10 centimeters and the length of Y X is 16 centimeters. Point G is the c

enter of the small circle. Point X is the center of the large circle. Points G, Y, and X are all on line segment GX. Marco wants to create a new circle using GX as a radius. What will be the area of Marco's new circle? 10Pi centimeters squared 16Pi centimeters squared 356Pi centimeters squared 676Pi centimeters squared

2 answers:

poizon [28]2 years ago

8 0

Answer:

its b

Step-by-step explanation:

damaskus [11]2 years ago

7 0

Answer:

The area will be $676\pi cm^{2}$

Step-by-step explanation:

I add a graph of the situation.

The first step is to find the length of the radius from the new circle.

The length will be the distance between the point ''G'' and ''X''.

This distance is equal to $10cm+16cm=26cm$

The new radius will be $26cm$

The area of a circle is equal to $\pi .R^{2}$ (I)

Where ''R'' is the radius.

Using the equation (I) :

$Area=\pi .(26cm)^{2}$

$Area=676\pi cm^{2}$

The area of the new circle will be $676\pi cm^{2}$

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The volume of the sphere is StartFraction 500 Over 3 EndFractionπ cubic units. A sphere has a radius of x units. What is the val

Tpy6a [65]

Answer:

(b) Radius of the sphere = 5 units.

Step-by-step explanation:

Here, given:

The volume of the sphere = $(\frac{500}{3})\pi$ cubic units ... (1)

Also, radius of the sphere = x units

Now, VOLUME OF SPHERE = $(\frac{4}{3}) \pi r^3$

So, volume of a sphere with radius x = $(\frac{4}{3}) \pi (x)^3$

But volume of sphere is given by (1).

$\implies (\frac{500}{3})\pi = (\frac{4}{3}) \pi (x)^3\\\implies (\frac{500}{3}) = (\frac{4}{3}) (x)^3\\\implies x^3 = (\frac{500}{3}) \times (\frac{3}{4}) = 125\\\implies x^ 3 = 125 = (5)^3\\\implies x = 5$

Hence, radius of the sphere = 5 units.

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On a coordinate plane, an absolute value graph starts at (0, 0) and goes up and to the left through (negative 4, 2). The functio

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Answer:

The function f(x)= –StartRoot negative x EndRoot is shown on the graph.

On a coordinate plane, an absolute value graph starts at (0, 0) and goes down and to the left through (negative 4, negative 2).

Step-by-step explanation:

The range of the graph is all real numbers greater than or equal to 0.

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Janet is designing a frame for a client. She wants to prove to her client that m∠AGE ≅ m∠CHE in her sketch. What is the missing

Amiraneli [1.4K]

Answer:

I believe the answer is transative property. I'm currently taking the test so I'm not 100% sure if I'm right yet.

Step-by-step explanation:

In step one, AGE and HGB are verticle angles since they are opposite of each other. then step two is alternate interior angles because they are interior angles opposite of each other, self-explanatory. the reason I choose the transitive property over the corresponding angles theorem is because the transitive property states: among A B and C, if A=B and B=c then A=C and in this case since AGE=HGB andHGB=CHE then AGE=CHE. I hope this helps and if you know the right answer just put it in the comments, if I'm wrong ofc.

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2 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

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