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OleMash [197]
3 years ago
9

I need help asap pls​

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:1/10

2/10

25/100

Step-by-step explanation:

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Find DY divided by DX by implicit differentiation y=sinxy 
amm1812

Answer:

Step-by-step explanation:

y = sin(xy)\\\frac{d}{dx}y=cos(xy)(\frac{d}{dx}(xy))\\  \frac{d}{dx}y = cos(xy)(y + x*\frac{d}{dx}y)\\ write \: \frac{d}{dx}y \: as\:y'\\y' = cos(xy)(y+xy')\\y' = ycos(xy) + xy'cos(xy) \\y'-xy'cos(xy) = ycos(xy)\\y'(1-xcos(xy)) = ycos(xy)\\\\y' = \frac{ycos(xy)}{1-xcos(xy)}

6 0
3 years ago
The following proof has an error. Find the error and choose the correct replacement.
Dvinal [7]

Answer:

opposite sides of a parallel are congruent.

Step-by-step explanation:

the symbol between AB and DC stands for "equal to", not parallel. boom roasted. (don't hate me if I'm wrong)

8 0
3 years ago
Deja has two baskets of berries. One of the baskets has 3 3/8 pounds of berries and the other basket has 2 7/8 pounds of berries
ElenaW [278]

Answer:

(b) The number of pounds of berries each person would receive is 1\frac{5}{8}pounds.

Step-by-step explanation:

The amount of berries in first basket = 3 3/8 pounds

Now, 3\frac{3}{8}  = 3+\frac{3}{8} = 3 + 0.375 = 3.375

So, the amount of berries in first basket = 3.375 pounds

The amount of berries in second basket = 2 7/8 pounds

Now, 2\frac{7}{8}  = 2+\frac{7}{8} = 2 + 0.875 = 2.875

So, the amount of berries in second basket = 2.875 pounds

Now, the total berries = Berries in ( First + Second)  basket

                                      = 3.375 pounds +  2.875 pounds  

                                    = 6.25 pounds

So, the number of pounds each person would have = \frac{\textrm{Total weight of viable berries}}{\textrm{4}}  = \frac{6.25}{4} = 1.5625

Now, 1.5625 = 1 + 0.5625  = 1 + \frac{5625}{10000}  = 1 + \frac{5}{8}  = 1\frac{5}{8}

So, the number of pounds of berries each person would receive is 1\frac{5}{8}pounds.

7 0
3 years ago
20 POINTS! HELP!!
astraxan [27]
The answer is C. 30.5% :)
5 0
3 years ago
Find three consecutive even integers such that 3 times the first is 26 less than twice the sum of the last two
Dimas [21]
Let's say our first integer is "a".

how to get the next consecutive EVEN integer?  well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".

and the next after that, will then be (a + 2) + 2, or "a + 4".

so those are are 3 integers, a           a + 2           a+4

notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.

2 + 2 is 4, or 8 + 2 is 10  some even ones

3 + 2 is 5, or 13 + 2 is 15, some odd ones

\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26}
\\\\\\
3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14
\\\\\\
0=a-14\implies 14=a

what are the other two consecutive integers?  well, a + 2 and a + 4.
4 0
3 years ago
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