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3 years ago

What happens to 35 when you multiply it by 100

Mathematics

2 answers:

MakcuM [25]3 years ago

8 0

When 35 is multiplied by 100, 35 is added 100 times, making 3,500

marusya05 [52]3 years ago

3 0

You times it by one and then add the two zeros so your answer is 3,500

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Use decimals and fractions in the same equation showing the Commutative Property. Repeat for the Associative Property.

Anna71 [15]

For Commutative Property of Addition

Let us take a decimal number $2.14$ and a fraction $\frac{5}{20}$

Now, according to the Commutative property of Addition:

For any two numbers $a$ and $b$ :

$a+b = b+a$

So, for $2.14$ and $\frac{5}{20}$

Let us add

$2.14+\frac{5}{20} = \frac{214}{100} +\frac{5}{20}$

$=\frac{214}{100} + \frac{5 \times 5}{20 \times 5} \\ \\=\frac{214}{100} + \frac{25}{100} \\ \\= 2.14 + 0.25 \\ \\ = 2.39$

Also,

$\frac{5}{20} + 2.14 = \frac{5}{20} + \frac{214}{100}$

$= \frac{5 \times 5}{20 \times 5} +\frac{214}{100} \\ \\= \frac{25}{100} + \frac{214}{100} \\ \\= 0.25 + 2.14 \\ \\ = 2.39$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Commutative Property of Addition is satisfied.

For Associated Property of Addition

Let us take two same decimal numbers $2.14$ , $7.25$ and a fraction $\frac{5}{20}$

Now, according to the Associated property of Addition:

For any three numbers $a$ , $b$ and $c$

$a + (b+c) =(a+b) + c$

So, for $2.14$ , $7.25$ and $\frac{5}{20}$

The Left hand side:

$a + (b+c)$

$2.14 + (7.25 + \frac{5}{20}) = 2.14 + (\frac{725}{100} + \frac{5 \times 5}{20\times 5})$

$= 2.14 + (\frac{725}{100} + \frac{25 }{100})$

$= 2.14 + (\frac{750}{100} )$

$= \frac{214}{100} + \frac{750}{100}$

$= \frac{964}{100}$

$=9.64$

The Right hand side:

$(a + b )+c$

$(2.14 + 7.25 )+ \frac{5}{20}= ( 9.39 ) + \frac{5 }{20}$

$= 9.39 + \frac{5 \times 5 }{20 \times 5}$

$= 9.39 + \frac{25}{100}$

$= 9.39 + 0.25$

$= 9.64$

Thus,

$(2.14 + (7.25 + \frac{5}{20} )= (2.14 + 7.25 )+ \frac{5}{20}$

Therefore, $2.14+\frac{5}{20} = \frac{5}{20} + 2.14$

Hence, Associative Property of Addition is satisfied.

8 0

2 years ago

Write a 3-digit number that is divisible by both 3 and 4. Explain how you know this number is divisible by 3 and 4.

Alexeev081 [22]

Answer:

let us assume the 3 digit number be 984

To check weather the number is divisible by 3 and 4 then when should know the divisibility rule of 3 and 4.

A number is said to be divisible by 3 when the sum of each digit of the number is divisible by 3. Example 9+8+4 = 21 is divisible by 3 hence the number is divisible by 3.

A number is said to be divisible by 4 when the last two digit is divisible by 4 like 84 is divisible by 4 hence 984 is divisible by 4.

5 0

2 years ago

Read 2 more answers

Pattie buys a handbag that costs 240 dollars and a scarf that costs 86 dollars. If sales tax is 7%, what is the total purchase p

Klio2033 [76]

Answer:

$348.82

Step-by-step explanation:

240 +86=326

326× 7%= 22.82

326+22.82= $348.82

7 0

2 years ago

Which transformations are needed to change the parent cosine function to y = 3 cosine (10 (x minus pi))?

Alekssandra [29.7K]

Answer:

See Explanation

Step-by-step explanation:

Given

New function: $y = 3 \cos(10 (x -\pi))$

We can assume the parent function to be:

$y = \cos (x)$

The new function can be represented as:

$y = A*\cos((\frac{2\pi}{B})(x-c))$

Where

A = Vertical stretch factor

B = Period

C = Right shift

By comparison:

$y = A*\cos((\frac{2\pi}{B})(x-c))$ to $y = 3 \cos(10 (x -\pi))$

$A = 3$

$c = \pi$

$\frac{2\pi}{B} = 10$

Solve for B

$B = \frac{2\pi}{10}$

$B = \frac{\pi}{5}$

Using the calculated values of $A,\ B\ and\ c.$ This implies that, the following transformations occur on the parent function:

Vertically stretched by $3$
Horizontally compressed by $\frac{\pi}{5}$
Right shifted by $\pi$

5 0

3 years ago

Read 2 more answers

Monique purchases a $5,100 dining room set. She can't afford to pay cash, so she uses the installment plan, which requires an 18

just olya [345]

Answer:

$918 is her down payment

Step-by-step explanation:

.18×5100

7 0

3 years ago