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viktelen [127]
3 years ago
15

How can i solve to find NO

Mathematics
1 answer:
Fittoniya [83]3 years ago
6 0

Answer:

3x+2+16=38

3x+18=38

3x=38-18

x=20/3

x=6.66

MN=3x+2

=3×6.66+2

=21.98

NO=38-21.98

=16.02

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Solve 2x^2=11x+6 by factoring
Ede4ka [16]

Answer:

(x-6) (2x+1)

Step-by-step explanation:

1) move everything over to the left side, so subtract 11x and 6.

2x^2 -11x - 6 = 0

2)multiply 2(a term) with -6 (c term)

x^2 -11x -12

3) factor and find what multiples to -12 and adds up to -11. in this case its -12 and positive 1

(x-12) (x+1)

4) divide -12 and 1 by the original a term (2)

(x-6) (x+1/2)

5) move the denominator of 2 over to the x.

(x-6) (2x+1)

5 0
4 years ago
How do i do these? (do at least one please)
meriva
32. 5^-2=1/25
5^-1=1/5
5^0=1
5^1=5
5^2=25
5^3=125

The domain are -2,-1,0,1,2,3,
The range are 1/25,1/5,1,5,25,125

So as the domain increases so does the range
3 0
3 years ago
How do you solve-<br> 10=4x
lana [24]

Answer:

x = 2 1/2

Step-by-step explanation:

4 0
3 years ago
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How does the position of a digit in a number relate to its value
Dafna1 [17]
The place value changes depending on where the # is. example 627 since the 7 is last its in the ones place. the two is in the tens and the 6 is in the hundreds
4 0
3 years ago
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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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