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Soloha48 [4]
3 years ago
10

6. The temperature of an iron block decreases from 85°C to 25°C. If the mass for

Mathematics
1 answer:
Fittoniya [83]3 years ago
4 0

Answer:

8280 cal

Step-by-step explanation:

Q=mc(delta)T

Q=1200(0.115)(-60)

Q=8280 cal

I used 1200 g instead of 1.2 kg

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What  is 7 x 1,829 in expanded form
Hoochie [10]
The answer:  7 * 1,829 =  " 12,803 " .
______________
<span>The following is the explanation—"in expanded form" — (as per the specfic instructions— within this very question—as to how to get the answer:
</span>____________________
Given:  7 * 1,829 = ?  ;   Find the solution; using "expanded form" :
____________
  (7 * 9 = 63 ) ; +

    (7 *20 = 140) ;  +

    (7 * 800 = 5,600) ; +
___________________________________________
    (7 * 1,000 = 7,000)  ; 
___________________________________________
Now, add the the values together to solve the problem:
___________________________________________
       →   7 * 1,829 = 63 + 140 + 5,600 + 7,000  ;
                           { =  203 + 5,600 + 7,000 } ;
                           { =  5,803  +  7,000 }  ;
                             =  12,803 ; which is the answer.
________________________________________________
Alternately, write out the steps as follows—using "expanded form":
________________________________________________
→ 7 * 1,829 = ?
________________________________________________
               →  7 * 1,829   =  (7*9) + (7*20) + (7*800) + (1,000) ; 
________________________________________________
     →   7 * 1,829  =   63 + 140 + 5,600 + 1,000 ;
                            { =  203 + 5,600 + 7,000 } ;
                            { =  5,803  +  7,000 }  ;
                             =  12,803 ;  which is the answer.
____
   → {Now, is our obtained answer:  "12,803" ;  the "correct answer"—to the problem:  " 7 * 1,829 " ;} ??

   → Let us check:   {Note:  " 7 * 1,829 " ;  is the same as:  ↔ " 1,829 * 7 " .}.

→  Using a calculator, does:  "7 * 1829 = ? 12,803" ?? ; Yes! ;
 →  &, for that matter; does:  " 1829 * 7 =? 12,803" ?? ;  Yes! .
_______
Furthermore, let us check, using the "traditional format" ; 
 →  Does:  "1,829 * 7 =?  12,803 ?? " ; 
________
{NB:  We are multiplying 2 (TWO) numbers together; & 1 (ONE) of these 2 [TWO] numbers is a "1-digit" ["single-digit"] number; & the "OTHER" multiplicand is a "multiple-digit" [specifically, a"4-digit"] number.}.
______
NB:  Yes; using a calculator is sufficient.  Below, I simply provide an alternate method to confirm whether our "obtained value" is correct.
_____
  →  Does:  "7 * 1,289 = ?  12,803" ?? ;
  →  Using the "traditional method"; let us check;  as follows:
_____
               ₅  ₂ ₆  
       →    1, 829 
  <span>      <u>     *        7         </u>                                                                </span>
             12   8 03 ;  
_____
So;  does:  "12,803 =? 12,803" ?? ;  YES!  
  → This "traditional method" shows that:  "7 * 1,829" ;  does, in fact, equal:  "12,803".
_____
{NB:   Explanation of the steps used in solving the aforementioned problem using the "traditional method"—just for clarification and confirmation} :
_____
→Start with:  "7*9= 63" ;  Write down the "3" & 'carry over' the "6" ; {Note the small-sized digit, "6"; written on top of the "2"; {commonly done—to keep track);

→Then;  "7*2 = 14" ; then add the "small digit 6"; to the "14" ; →"14+6 =20" ;
Write down the "0" ; & 'carry over' the "2" ; {Note the "small-sized digit, "2"; written over the "8"; (commonly done—to keep track);

→ Then; "7*8 = 56" ;  then add the "small digit 2"; to the "56"; → "56+2 = 58" ; Write down the "8" ; & 'carry over' the "5" ; {Note the "small-sized digit", "5" ; written over the "1" ; (commonly done—to keep track);

→Then;  "7*1 = 7" ; then add the "small digit 5"; to the "7" ; → "7+5 = 12" ; Write down the "12" ; in its entirety—since are no digits left [in the multiplicand, "1,829"] ; to "carry over". 
____
We get:  "12,803" ;  which =? "12,803" ?? ;→Yes!
____
I hope my explanation of how to solve "7 * 1,829" ; using the "expanded form" is helpful.  Also, i hope my explanation—albeit lengthy— of confirming that  [<em>our</em>]  "correctly obtained value"—which is:  "12,803"— is of some help.
__
7 0
3 years ago
Please help me thank you
Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

3 0
3 years ago
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