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3 years ago

Help pleaseeeeeeee!!!!

Mathematics

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

there's an app called mathpapa and if you type in the equation it should answer it for you. or just go to the website. that's how I got passed algebra. GEOMETRY is hard

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HELP PLZ!!!i cant figure this out and dont worry about what else im typin it wont let me continue without having 20 word typed s

morpeh [17]

Answer:

The length is 42 inches and the width is 14 inches

Step-by-step explanation:

You know the width is 1/3 of the length, and the length is basically the longer side of a rectangle. The width is y-4, so multiply using distributive property to get 3y-12.

You now know that the 2y+6 is equal to 3y-12, so you add 12 to both sides using additive property of addition to get 2y+18=3y.

Then you subtract 2y from both sides to get the unit value of y.

Y is equal to 18.

Then you can plug 18 into the y's to get

2(18)+6 which is 36+6 = 42

18-4 = 14

There you go.

The length is the longer side which is 42 inches and the width is 14 inches

Hope this helps!

6 0

3 years ago

Use complete sentences to explain how to easily find the expansion of (x + 2)9.

Vaselesa [24]

The easiest way to find the expansion of the expression given would e to distribute. To distribute, you have to multiply each of the terms in the parentheses by 9. The expansion of the expression would be 9x + 18.

9 (x + 2)
9 · x = 9x
9 · 2 = 18
9x + 18

Hope this helps. Good luck! :)

4 0

3 years ago

The sum of 11.6 and 4.1 by rounding each number to the nearest whole.

Karo-lina-s [1.5K]

11.6 rounded to the nearest whole is 12 and 4.1 rounded to the nearest whole is 4 (rule is if the decimal [or number in general] is 4 or lower, the whole stays the same and if the number is 5 or higher, the whole goes up by one). 12+4=16

3 0

2 years ago

What is the value of the six in 106,534

aleksley [76]

6,000 because its in the thousandth and so since its a six in the thousandths its 6,000 remember the comma

8 0

3 years ago

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers

love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = $7 \times 7 \times 6$ = 294

Part B:

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $6 \times 6 \times 5$ = 180

Part C)

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $3 \times 5 \times 5$ = 75

Part d)

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = $3 \times 7 \times 7$ = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = $1 \times 3 \times 7$ = 21

Total number of required ways = 147 + 21 = 168

Part e)

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = $3 \times 6 \times 5$ = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = $1 \times 3 \times 5$ = 15

Total number of required ways = 90 + 15 = 105

7 0

3 years ago