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torisob [31]
2 years ago
6

A triangle with base 6.2 cm and height 4.8 cm

Mathematics
1 answer:
Eva8 [605]2 years ago
5 0

Answer:

14.88

A = \frac{h_bb}{2}

    = 4.8 × 6.2 / 2

     = 14.88

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I cant seem to answer this question
ale4655 [162]

Answer:

yes

Step-by-step explanation:

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What is 75-5(2x6) ?
otez555 [7]
75-5(2×6) 75-5×12× -60×+75
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Which is the correct form of the quadratic formula? A) x = b ± b2 - 4ac a B) x = b ± b2 - 4ac 2a C) x = - b ± b2 - 4ac a D) x =
DochEvi [55]

That's pretty ill-formatted but the answer is D

ax^2 + bx + c = 0

has roots given by

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Bonus quadratic formulas:

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x^2 - 2bx + c \textrm{ has zeros at }x=b \pm \sqrt{b^2-c}

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4 0
4 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

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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}
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Answer:

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Step-by-step explanation:

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