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lukranit [14]
3 years ago
14

The sum of the digits of a two-digit number is 5. If nine is subtracted from the number, the digits will be reversed. Find the n

umber. the answer to that is 32. If you replace the tens digit with x, then what will the equation be?
Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer: 32

Step-by-step explanation:

Let x + y = 5

x = 5 - y....................(1)

10x + y -9 = 10y + x..... . (2)

Substituting (1) in (2)

10(5-y) + y - 9 = 10y + (5-y)

50-10y + y - 9 = 10y + 5 - y

41 - 9y = 9y + 5

-9y - 9y = 5-41

-18y = - 36

y = 36/2 = 2

So, x = 5-2 = 3

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5-2.1=? 5-2.17=? 5-2 7/8 solve make sure to re-check your self before answering or no forty points
frozen [14]

Answer: 5-2.1= 2.9

5-2.17= 2.83

5-2 7/8= 2.125

6 0
3 years ago
-2x + 9x<br> What is the answer
Blababa [14]

Answer: 7x

<u>Remove "x" and add</u>

-2 + 9 = 7

<u>Add "x" back</u>

7 -----> 7x

Note: When adding/subtracting problems like these remove "x" and then add/subtract normaly. When you get your answer put x back.

7 0
3 years ago
Read 2 more answers
Which expression is equivalent??? help!
dexar [7]

Answer:

The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

Step-by-step explanation:

Given:

\sqrt[3]{256x^{10}y^{7} }

Solution:

We will see first what is Cube rooting.

\sqrt[3]{x^{3}} = x

Law of Indices

(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}

Now, applying above property we get

\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y   )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})

∴ The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

5 0
3 years ago
(a)
Allisa [31]

Answer:

320,000

Step-by-step explanation:

4 0
3 years ago
Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
Marizza181 [45]

If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams

The half life in years = 24100

Consider the quantity of the radio active isotope remaining

y = Ce^{kt}

When t = 1000 the y = 1.2

y = C/2 when t = 1599

Substitute the values in the equation

C/2 = Ce^{24100k}

Cancel the C in both side

1/2 = e^{24100k}

Here we have to apply ln to eliminate the e terms

ln (1/2) = 24100k

k = ln(1/2) / 24100

k = -2.87× 10^-5

To find the initial value we have to substitute the value of k and y in the equation

1.2 = Ce^{1000 ×  -2.87× 10^-5}

C = 1.2 / e^(-0.0287)

C = 2.16 gram

Hence, the initial quantity of the radioactive isotope is 2.16 gram

Learn more about half life here

brainly.com/question/4318844

#SPJ4

7 0
1 year ago
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