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3 years ago

Please answer correctly and no links

Mathematics

1 answer:

denis23 [38]3 years ago

4 0

Answer:

$\frac{1}{5}$ (m - 100)

Step-by-step explanation:

If you multiply out the bracket, you'll get $\frac{1}{5}$ m - 20

Try multiplying out the brackets in the other answers and you'll get something different

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What is

d1i1m1o1n [39]

Answer:

9y-3y=6y is equivalent to 9y – 3y.

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3 years ago

(Q5) Decide if the function is an exponential function. If it is, state the initial value and the base. y=3.1^x

ycow [4]

Answer:

Step-by-step explanation:

Exponential equation takes the form $y=ab^x$ where

a is the initial value ( a ≠ 0), and

b is the base ( b ≠ 1)

The equation given can be written as $y=3.1^x\\y=1*3.1^x$ . Thus, it is an exponential equation.

So a = 1 and b = 3.1

Thus we can say that the initial value = 1 and the base is 3.3

6 0

3 years ago

You roll a 100 sided dice where you get paid the number showing on the top face. How much money would someone pay to play this g

Dahasolnce [82]

Answer:

$50

Step-by-step explanation:

$50 would be the payment because it would be 50/50 if you get lower than 50 or higher than 50.

7 0

3 years ago

If you divided 20acres of land into 1\4 acre section how !any sections doses it divied to

LuckyWell [14K]

If you divide 20 acres into 1/4 sections you'll get
20 ÷ 1/4 sections. When you divide fractions, you just multiply the reciprocal.

20 ÷ 1/4 =
20 × 4/1 = 80 sections

5 0

4 years ago

Amy's grandmother gave her 3 identical chocolate chip cookies and 4 identical sugar cookies. In how many different orders can Am

Dmitrij [34]

Answer:

chocolate chip cookie first: 15 ways

chocolate chip cookie last: 15 ways

chocolate chip cookie first and last: 5 ways

Step-by-step explanation:

There is a total of 7 cookies.

Is she eats a chocolate cookie first, she will have 2 chocolate cookies and 4 sugar cookies (6 in total).

So, to find the number of different orders that Amy can eat the remaining cookies, we just need to calculate a combination of 6 choose 2 (that is, the number of ways we can put the 2 chocolate cookies among the 6 cookies) or a combination of 6 choose 4 (same thing, but for the sugar cookies):

C(6,2) = 6! / (2! * 4!) = 6 * 5 / 2 = 15 ways

Is she eats a chocolate cookie last, she will have 2 chocolate cookies and 4 sugar cookies (6 in total), so the problem is solved again with a combination of 6 choose 2:

C(6,2) = 15 ways

Is she eats a chocolate cookie first and last, she will have just 1 chocolate cookie left and 4 sugar cookies (5 in total), so the problem is solved with a combination of 5 choose 1 or a combination of 5 choose 4:

C(5,1) = 5! / (1! * 4!) = 5 ways

6 0

3 years ago