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bearhunter [10]
3 years ago
6

If I think of a number halve it and subtract 2 results is 10.write equations and solve it please

Mathematics
2 answers:
gayaneshka [121]3 years ago
6 0

Answer:

24

Step-by-step explanation:

let 'x' = number

x/2 - 2 = 10

add 2 to each side to get:

x/2 = 12

cross-multiply to get:

x = 24

grin007 [14]3 years ago
5 0

Answer:

  • The equation is (x/2) - 2 = 10
  • The equation solves to x = 24

============================================================

Explanation:

x = the number you think of

x/2 = the result after taking half of that number

(x/2) - 2 = subtract 2 from the previous result

(x/2) - 2 = 10 is the equation to solve

Solving said equation gets us...

(x/2) - 2 = 10

x/2 = 10+2 ... adding 2 to both sides

x/2 = 12

x = 12*2 ..... multiplying both sides by 2

x = 24

Notice how we basically undo everything applied to x. First we divided by 2, then we subtracted 2. So we follow that order in reverse to undo the "minus 2" (ie add 2 to both sides). Then we undo the division to multiply both sides by 2 to end up with the answer x = 24

-----------------

Let's check that answer:

x = 24

x/2 = 24/2 = 12 ... result after taking half of x

(x/2) - 2 = (12)-2 = 10 ... result after subtracting two

We end up with 10 as your teacher mentioned, so it confirms that x = 24 is the answer.

------------------

Here's another way to check:

(x/2) - 2 = 10

(24/2) - 2 = 10

(12) - 2 = 10

10 = 10

We end up with the same thing on both sides, so x = 24 is confirmed.

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x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

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AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

(x+5)(x-3) >= 0

This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

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)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

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Answer:

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Answer:

  • Solution of equation ( q ) = <u>1</u><u>6</u>

Step-by-step explanation:

In this question we have given an equation that is <u>3 </u><u>(</u><u> </u><u>q </u><u>-</u><u> </u><u>7</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u>and we have asked to solve this equation that means to find the value of <u> </u><u>q</u><u> </u><u>.</u>

<u>Solution : -</u>

\quad \: \longmapsto \:  3(q - 7) = 27

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis :

\quad \: \longmapsto \:3q - 21 = 27

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 21 on both sides :

\quad \: \longmapsto \:3q -  \cancel{ 21} +  \cancel{21} = 27  +  21

On further calculations we get :

\quad \: \longmapsto \:3q = 48

<u>Step </u><u>3 </u><u>:</u> Dividing by 3 from both sides :

\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}}  =  \cancel {\frac{48}{3} }

On further calculations we get :

\quad \: \longmapsto \:   \pink{\boxed{\frak{q = 16}}}

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<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are very our answer by substituting value of q in the given equation . So ,

  • 3 ( q - 7 ) = 27

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  • 3 ( 9 ) = 27

  • 27 = 27

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> </u><u>our </u><u>solution</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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