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dybincka [34]
3 years ago
8

Division problems to solve for 4th grade

Mathematics
1 answer:
Scrat [10]3 years ago
4 0
10 ÷ 2
27 ÷ 3
72 ÷ 8
84 ÷ 21 (can be difficult for some)
24 ÷ 8
69 ÷ 3
56 ÷ 7
39 ÷ 13
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23/40 as a percentage
Leviafan [203]
23/40* 100%= 57.5%
Hope this helps~

7 0
3 years ago
Order the numbers from least to greatest.<br> 17.2.3
nordsb [41]
The order of these numbers is 2, to 3, to 17. hope this helps :))
5 0
3 years ago
Pleaseeeee helpppp!!!
Gnesinka [82]

Answer:

540

Step-by-step explanation:

No matter what the shape, the total amount of degrees in the interior angles in a pentagon is 540. Hope this helps!

4 0
3 years ago
Verify the identitiy:
Advocard [28]

Answer:

\frac{sinx}{1-cos x}     =         cosecx  +  cot x

Step-by-step explanation:

To verify the identity:

sinx/1-cosx = cscx + cotx

we will follow the steps below;

We will take just the left-hand side and work it out to see if it is equal to the right-hand side

sinx/1-cosx

Multiply the numerator and denominator by 1 + cosx

That is;

\frac{sinx}{1-cos x}     =    \frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}

open the parenthesis on the right-hand side of the equation at the numerator and the denominator

sinx(1+cosx) = sinx + sinx cosx

(1-cosx)(1+cosx) = 1 - cos²x

Hence

\frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}     =  \frac{sinx + sinx cosx}{1-cos^{2}x }

But 1- cos²x  = sin²x

Hence we will replace  1- cos²x  by  sin²x

   \frac{sinx}{1-cos x}    =       \frac{sinx(1+cosx)}{(1-cosx)(1+cosx)}     =  \frac{sinx + sinx cosx}{1-cos^{2}x }   =  \frac{sinx+sinxcosx}{sin^{2}x }

                             

                                  =\frac{sinx}{sin^{2}x }   +   \frac{sinxcosx}{sin^{2}x }

                                   

                                   =\frac{1}{sinx}   +   \frac{cosx}{sinx}

             

                                   =cosecx  +  cot x

\frac{sinx}{1-cos x}     =         cosecx  +  cot x

Note that;

\frac{1}{sinx}  = cosecx                        

         

 \frac{cosx}{sinx}   =       cot x

                                     

6 0
4 years ago
Subtract the cube root of the product of h and 3k from the square of the sum of p and q​
ehidna [41]

Answer:

(p + q)²            -          ∛(h·3k)   or   (p + q)² - ∛(h·3k)  

Step-by-step explanation:

Cube root of x:  ∛x

Product of h and 3k:  h·3k

Sum of p and q:  p + q

*****************************

From (p + q)²      subtract ∛(h·3k)      This becomes, symbolically:

=>       (p + q)²            -          ∛(h·3k)

3 0
3 years ago
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