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3 years ago

Does anyone know this

Mathematics

1 answer:

Katen [24]3 years ago

5 0

Answer:

22 cm

Step-by-step explanation:

All angles of a triangle must add up to 180,

So we just take angle ACB and angle EFG, add them and then subtract the sum from 180

(angle ACB + angle EFG) - 180 = angle GEF

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apply the function transformation of g(x)=f(x-2)+4 to f(x)=x^3-8. describe the transformation and find the final form of the fun

Fittoniya [83]

Answer:

g(x) = (x - 2)^3 - 4

Step-by-step explanation:

first, the function f is shifted to the right by 2 units and then shifted up by 4 units

4 0

4 years ago

How do I find the volume of this?

lorasvet [3.4K]

Answer:

(64+320π) cm^3

Step-by-step explanation:

The volume of a cube is the length*width*height, so

the volume of this cube is 4*4*4=64 cm^3

The volume of a cylinder is the base*height

The base of a cylinder is the area of a circle, which is π*the radius of the circle squared

The diameter of the circle base is 16, and the radius is half the diameter, so it is 16/2=8.

The area of the base of the cylinder is π8^2=64π

Now, multiply the base by the height, 5

64π*5=320πcm^3

We now have the volume of the cube and the volume of the cylinder, so now we just add them:

(64+320π) cm^3

7 0

3 years ago

A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow

Mrrafil [7]

Answer:

$\displaystyle \frac{54}{5405}$ .

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

$\displaystyle \left(50\atop 5\right) = 2,118,760$ .

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

$\displaystyle \left( 28\atop 2\right) = 378$ .

Number of ways to choose 3 green candies out of a batch of 8:

$\displaystyle \left(8\atop 3\right)=56$ .

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

$\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168$ .