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3 years ago

(6x^1) ^2 Please show all work

1 answer:

6 0

1. Apply a^1= a , rule.
x^1 = x
=(6x)^2

2. Then apply exponent rule. (a*b)=a^n b^n
=6^2 x^2

3.6^2 = 36 , because 6 x 6 = 36
Now, apply x^2
=36x^2

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A jar contains 45 red candies and 60 black candies. Suppose a candy is selected at random. What are the odds against selecting a

uysha [10]

Answer:

3:4.

Step-by-step explanation:

To work this out we need to find the highest multiple of 45 and 60.

15 is the largest number that goes into both of them so what we are going to do now is divide both number by 15.

45 divided by 15 = 3

60 divided by 15 = 4

Therefore the odds of selecting a red candy is 3:4.

Hope that helps. x

7 0

2 years ago

[(8*9)] - (6*7)] -15

maks197457 [2]

Answer:

Step-by-step explanation:

Simplify the following:

8×9 - 6×7 - 15

8×9 = 72:

72 - 6×7 - 15

-6×7 = -42:

72 + -42 - 15

72 - 42 - 15 = 72 - (42 + 15):

72 - (42 + 15)

| 4 | 2

+ | 1 | 5

| 5 | 7:

72 - 57

| 6 | 12

| 7 | 2

- | 5 | 7

| 1 | 5:

Answer: 15

3 0

2 years ago

Read 2 more answers

Gina wants to ship 3 books that weigh 2 7/16 pounds, 1 7/8 pounds and 1/2 pound. the maximum weight she can ship is 6 pounds. ho

pshichka [43]

You have to find a common denominator 1st then add. So 16 is the common denominator the problem should be 2 7/16+1 14/16+8-16= 3 29/16 then you need to divide 29/16 you will have 4 13/16

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3 years ago

Read 2 more answers

What is the measure of XYZ 36 Degrees <br> Please help

exis [7]

B is the correct answer

good luck

8 0

3 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola