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nikitadnepr [17]
3 years ago
5

Find all the second order partial derivatives of g (x comma y )equalsx Superscript 4 Baseline y plus 5 sine (y )plus 4 y cosine

(x ).

Mathematics
1 answer:
Dovator [93]3 years ago
6 0

Answer:

Step-by-step explanation:

Check attachment for solution

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Please help if you can!!!!
storchak [24]

Answer:Well, I don't know what you got so I can't tell you if it is right.

If it works in both equations, it depends of whether your equations are set up correctly.

Here is how I would do this problem.

Let x = no. of hot dogs,y = number of sodas.

First equation is just about the number of things.

x + y = 15

Second equation is about the cost of things.

1.5 x + .75 y = 18

solve x+y = 15 for y  y = 15-x    substitute into second equation

1.5x + .75(15 - x) = 18    

You should get the correct answer for number of hot dogs if you solve this correctly.  Put your answer in the x + y =15 equation to get y.  Then put both x and y into the cost equation and check your answer.

Hope this helps.

Step-by-step explanation:

5 0
3 years ago
Refer to the figure below to complete the following item. Given: Quadrilateral ROSE is trapezoid with median If MN = 24 and ES =
STatiana [176]
<span>18 is you answer. Hope this helps.</span>
4 0
3 years ago
Read 2 more answers
Which function is equivalent to F(x)=-4(x+7)^2-6?
Gemiola [76]

\huge\mathfrak\orange{answer}

Simplify this quadratic function given in vertex form into standard form by simplifying the exponents and adding like terms.

f(x) = \red{-4(x+7)^2 - 6}

f(x) = -4(x^2 + 14x + 49) - 6

f(x) = -4x² - 64x - 196 - 6

\underline{f(x) = -4x² - 64x - 202}

(good luck :) and mark me brainliest if you're satisfied with my answer)

7 0
3 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
Ksju [112]

Answer:converge at I=\frac{1}{3}

Step-by-step explanation:

Given

Improper Integral I is given as

I=\int^{\infty}_{3}\frac{1}{x^2}dx

integration of \frac{1}{x^2}  is  -\frac{1}{x}

I=\left [ -\frac{1}{x}\right ]^{\infty}_3

substituting value

I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]

I=-\left [ 0-\frac{1}{3}\right ]

I=\frac{1}{3}

so the value of integral converges at \frac{1}{3}

8 0
4 years ago
100 points if you guess what number im thinking.1-5
posledela

Answer:

Step-by-step explanation:

5

3 0
4 years ago
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