The perimeter of a triangle is 19z-7. One side is 5z-9 and one side is 8z+3. What is the remaining side ?

Consider all seven-digit numbers that can be created from the digits 0-9 where the first and last digits must be odd and no digi

Yuliya22 [10]

Answer:

1/4 or 0.25

Step-by-step explanation:

The total possibilities of any 7 digit number using 0-9 is :

9×10×10×10×10×10×10=9000000

To work out the total possibilities in this question :

We look at the conditions :

The first digit can only be 5 numbers :

1 , 3 , 5 , 7 , 9

Now we subtract 5 from 9 :

9-5 = 4

Since no repeats for 2 , 3 , 4, 5, 6:

9 , 8 , 7 , 6 , 5,

5 possibilities for the last digit :

Total possibilities for this code :

4 × 9 × 8 × 7 × 6 × 5 × 5 = 302400

If it begins with 5 that is only 1 possibility for the first digit

1 × 9 × 8 × 7 × 6 × 5 × 5 = 75600

Now we make a fraction :

75600÷302400

Dividing top and bottom by 75600 gives you 1/4 or 0.25

Hope this helped and have a good day

7 0

2 years ago

Read 2 more answers

Giving right answer a branleist

Kobotan [32]

Answer:

(a) 860, (b) 860, (c) 860 and 186

Step-by-step explanation:

(a) 860

860 ends with a zero or a five, 186 and 863 do not.

(b) 860

860 ends with a zero, 186 and 863 do not.

(c) 860, 186

860, the 0 is even. 186, the 6 is even. 863, the 3 is not even.

7 0

2 years ago

Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

Elan Coil [88]

Answer:

(40/3)*pi

Step-by-step explanation:

lets first find the area of the little circle

A=pi*r^2

A=pi*3^2

A=9pi

Now, lets find the area of the big circle

A=pi*r^2

A=pi*(3+4)^2

A=pi*7^2

A=49pi

now lets subtract the area of the little circle from the area of the big circle

49pi-9pi=40pi, now we found the area of the "bagel)

lets find what portion of the bagel is the shaded region

120/360=1/3

now lets multiply the bagel area by the fraction

40pi*(1/3)=

(40/3)*pi or (40*pi)/3

4 0

3 years ago

A radioactive substance decreases in the amount of grams by one-third each year. If the starting amount of the

katovenus [111]

Answer:

The sequence is geometric. The recursive formula is $a_{n}=2/3a_{n-1}$

Step-by-step explanation:

In order to solve this problem, you have to calculate the amount of the substance left after the end of each year to obtain a sequence and then you have to determine if the sequence is arithmetic or geometric.

The substance decreases by one-third each year, therefore:

After 1 year:

$1452-\frac{1}{3}(1452)$

Using 1452 as a common factor and solving the fraction:

$1452(1-\frac{1}{3})=1452(\frac{2}{3})=968$

You can notice that in general, after each year the amount of grams is the initial amount of the year multiplied by 2/3

After 2 years:

$968(\frac{2}{3})=\frac{1936}{3}$

After 3 years:

$\frac{1936}{3}(\frac{2}{3})=\frac{3872}{9}$

The sequence is:

1452,968,1936/3,3872/9....

In order to determine if the sequence is geometric, you have to calculate the ratio of two consecutive terms and see if the ratio is the same for all two consecutive terms. The ratio is obtained by dividing a term by the previous term.

The sequence is arithmetic if the difference of two consecutive terms is the same for all two consecutive terms.

-Calculating the ratio:

For the first and second terms:

968/1452=2/3

For the second and third terms:

1936/3 ÷ 968 = 2/3

In conclussion, the sequence is geometric because the ratio is common.

The recursive formula of a geometric sequence is given by:

$a_{n}=ra_{n-1}$

where an is the nth term, r is the common ratio and an-1 is the previous term.

In this case, r=2/3

7 0

3 years ago

Caroline has some dimes and some quarters. She has a maximum of 15 coins worth at least $2.85 combined. If Caroline has 3 dimes,

gayaneshka [121]

Answer:

11, 12.

Step-by-step explanation:

Let q represent number of quarters.

We have been given that Caroline has a maximum of 15 coins worth at least $2.85 combined. We are also told that Caroline has 3 dimes. This means that total coins are less than or equal to 15.

We can represent this information in an inequality as:

$q+3\leq 15...(1)$