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3 years ago

Ellen buys 4 bags of oranges. There are 6 oranges in each bag. How many oranges does Ellen buy?

Mathematics

2 answers:

ser-zykov [4K]3 years ago

8 0

Answer:

24 oranges.

Step-by-step explanation:

Multiply 6 oranges by 4 bags.

6 x 4 = 24.

Therefore, Ellen buys 24 oranges.

Hitman42 [59]3 years ago

4 0

Answer:

Step-by-step explanation:

Total bags of orange = 4bags

There are 6 orange in a bag

Total orange in 4bags = 4 × 6

= 24 Oranges

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3 years ago

A family pass to the amusement park costs $54 Using the Distributive Property,write an expression that can be used to fine the c

snow_lady [41]

X represents the amount of children's tickets
y represents the amount of adult's tickets

8(x+y) = 54

7 0

2 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

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

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

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

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

a companys product development division has 60 employees. Of these, 1/3 are engineers. If 40% of engineers are moved to another

Musya8 [376]

Answer: There will be 12 engineers remaining in the product development department.

Step-by-step explanation:

So the givens are:

60 employees and 1/3 are engineers and if 40% of engineers are moved to another division and how many engineers will remain in the product development department?

So what we need to do:

60 * 1/3 = 60 / 3 = 20

There are 20 engineers.

20 * 40% = 8

There are 8 engineers that are g<u>oing to be moved to another division</u>.

20 - 8 = 12

12 engineers will <u>remain</u> in the development department.

Hence, your answer.

6 0

3 years ago

How many groups of 25 are there in 500

Elis [28]

500÷25

they are 20 groups

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3 years ago

Read 2 more answers