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yaroslaw [1]
3 years ago
13

Find the square root of 121 using the method of repeated subtraction​

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0
Your answer is 11. This has to be 20 characters long in order to submit.
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Marco Walked 2/100 kilometers to the park. Then he walked 3/10 kilometer to the library. What fraction of a kilometer did he wal
pickupchik [31]
57/100 is the answer :)
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Juan has grown 2 inches since last summer, but he's still not tall enough to get on the Crazy Carts ride. The minimum height for
Ksenya-84 [330]

Answer:

x + 2 < 42

Step-by-step explanation:

Minimum height for Crazy cart = 42 inches

x = height last summer

Number if inches grown since last summer = 2

Height now = x + 2 is still not yet up to 42 inches because he is not yet tall enough to get on crazy carts ride.

Representing as an inequality :

x + 2 < 42

8 0
3 years ago
HELP- 11 points!
kvv77 [185]
(x)+(x+1)+(x+2)=-36
3x+3=-36
3x=-39
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The numbers are -13, -12, -11
5 0
3 years ago
Help i hate .math..........
RideAnS [48]

6 divided by \frac{8}{9} is the same as 6/1 multiplied by 9/8, which gives us 7 as the whole answer. therefore, I think number 1 and 4 are the correct selections

3 0
3 years ago
Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy
nadya68 [22]
ANSWER TO QUESTION 1

The given function is

f(x,y)=2 {e}^{xy}

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

f_{x} = 2y {e}^{xy}

and the second derivative with respect to x is,

f_{xx} = 2 {y}^{2} {e}^{xy}

ANSWER TO QUESTION 2

The given function is

f(x,y)=2 {e}^{xy}

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

f_{y} = 2x{e}^{xy}

and the second derivative with respect to y is

f_{yy} = 2 {x}^{2} {e}^{xy}

ANSWER TO QUESTION 3

Our first mixed partial is

f_{xy}

We need to differentiate
f_{x} = 2y {e}^{xy}
again. But this time with respect to y.

Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,

f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'

f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}

ANSWER TO QUESTION 4

The second mixed partial is

f_{yx}

We need to differentiate
f_{y} = 2x{e}^{xy}

again. But this time with respect to x.

Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,


f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'

f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}

Hence,

f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}
4 0
3 years ago
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