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lorasvet [3.4K]

3 years ago

6

What is sum or difference for 2/5 2/3

1 answer:

Kisachek [45]3 years ago

5 0

6/15 + 10/15 = 16/15 = 1 1/15

6/15 - 10/15 = -4/15

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What is the snswer to 3a+2b x 7c

Ad libitum [116K]

I am pretty sure your answer is this because 3+2=5 and 5*7=35 abc

3 0

3 years ago

Read 2 more answers

Crispy Clover, a popular vegetarian restaurant, introduced a new menu that has 20%, percent more dishes than the previous menu.

Elden [556K]

Correct answers are $D + D(\frac{1}{5})$ and $1.2D$ i.e. Options A & D

<u>Step-by-step explanation:</u>

Options for this questions are:

A)D+1/5D

B)D+20

C)D+20D

D)1.2D choose two answers.

We have , Crispy Clover, a popular vegetarian restaurant, introduced a new menu that has 20%, percent more dishes than the previous menu. The previous menu had D dishes.We need to find expressions could represent how many dishes Crispy Clover's new menu has . Let's find out:

Initially , Number of dishes are D . Then 20% more dishes are there in menu

⇒ $\frac{D(20)}{100}$

So , Total Number of Dishes are

⇒ $D + \frac{20d}{100}$

⇒ $D + \frac{20d}{20(5)}$

⇒ $D + D(\frac{1}{5})$

⇒ $D + D(0.2)$

⇒ $1.2D$

Therefore, Correct answers are $D + D(\frac{1}{5})$ and $1.2D$ i.e. Options A & D .

4 0

3 years ago

Can someone please help me with this?

jonny [76]

I need to know if you where are you gonna is the day you

4 0

3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

3 0

3 years ago

NEED HELP ASAP PLEASE

vovikov84 [41]

Answer:

The fourth one I believe

Step-by-step explanation:

3 0

3 years ago