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3 years ago

I dont understand this

Mathematics

1 answer:

mylen [45]3 years ago

3 0

Answer:

ye me too i cant answer thay

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Elmer borrowed $3150 from his brother Julio to pay for books and tuition. He agreed to repay

Vladimir79 [104]

Answer: $3,276

Step-by-step explanation:

To find the amount of interest needed to be paid, you needed to multiply the amount of money by your interesate rate.

= 3150 x 0.04

= 126

So, we add this extra amount to be paid to the exisiting total

= 3150 + 126

= $3,276

3 0

2 years ago

X over 2 -8=3<br> How do I find the answer if I do not get it plz help

Klio2033 [76]

$\dfrac{x}{2}-8=3\\
\dfrac{x}{2}=11\\
x=22$

7 0

3 years ago

A student has two test scores. the difference between the scores is 44 and the mean, or average, of the scores is 8484. what a

GrogVix [38]

It is 22 above and 22 below, so the scores are 8462 and 8506

7 0

4 years ago

I need help. Find the value of each variable.

azamat

Answer:

x=8, y=4 $\sqrt{3}$

Look up the 30-60-90 triangle rule

8 0

3 years ago

Read 2 more answers

Local versus absolute extrema. If you recall from single-variable calculus (calculus I), if a function has only one critical poi

stich3 [128]

Answer:

Step-by-step explanation:

Given that:

$f(x,y) = e^{3x} + y^3 - 3ye^x \\ \\ \implies \dfrac{\partial f}{\partial x} = 0 = 3e^{3x} -3y e^x = 0 \\ \\ e^{2x}= y \\ \\ \\ \implies \dfrac{\partial f}{\partial y } = 0 = 3y^2 -3e^x = 0 \\ \\ y^2 = e^x$

$\text{Now; to determine the critical point:}$ - $f_x = 0 ; \ \ \ \ \ f_y =0$

$\implies e^{2x} = y^4 = y \\ \\ \implies y = 0 \& y =1 \\ \\ since y \ne 0 , \ \ y = 1, \ \ x= 0\\\text{Hence, the only possible critical point= }(0,1)$

$\delta = f_xx, s = f_{xy}, t = f_{yy} \\ \\ . \ \ \ \ \ \ \ \ D = rt-s^2 \\ \\ i) Suppose D >0 ,\ \ \ r> 0 \ \text{then f is minima} \\ \\ ii) Suppose \ D >0 ,\ \ \ r< 0 \ \text{then f is mixima} \\ \\ iii) \text{Suppose D} < 0 \text{, then f is a saddle point} \\ \\ iv) Suppose \ D = 0 \ \ No \ conclusion$

$Thus \ at (0,1) \\ \\ \delta = f_{xx} = ge^{3x}\implies \delta (0,1) = 6 \\ \\ S = f_{xy} = -3e^x \\ \\ \implies S_{(0,1)} = -3 \\ \\ t = f_{yy} = 6y \\ \\$

$\implies t_{0,1} = 6$

$Now; D = rt - s^2 \\ \\ = (6)(6) -(-3)^2$

$= 36 - 9 \\ \\ = 27 > 0 \\ \\ r>0$

$\text{Hence, the critical point} \ (0,1) \ \text{appears to be the local minima}$

$\text{Suppose we chose x = 0 and y = -3.4} \\ \\ \text{Then, we have:} \\ \\ f(0,-3.4) = 1+ (-3.4)^3 + 3(3.4) \\ \\ = -28.104 < -1$

$\text{However, if f (0,1) = 1 +1 -3 = -1 \\ \\ f(0,-3.4) = -28.104} < -1} \\ \\ \text{This explains that} -1 \text{is not an absolute minimum value of f(x,y)}$

7 0

3 years ago