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3 years ago

FIRST ASWER WILL GET BRAINLIEST -2.45-(-3.9)

Mathematics

2 answers:

Rama09 [41]3 years ago

7 0

Answer:

1.45

Step-by-step explanation:

Lelechka [254]3 years ago

7 0

Answer:

1.45

Step-by-step explanation:

-2.45-(-3.9) :/

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Hi guys best answer gets a brainlist !!!!!!! <br> please view image below

saw5 [17]

Answer:

Step-by-step explanation:

since the linear function intersects the x-axis at -1, the coordinate for the x-intercept is (-1,0). Since the line also intersects the y-axis at 3, the pint of intersection is (0,3)

Hope I helped UwU

3 0

3 years ago

Y = 6x + 11<br> 2х - Зу = 7

irakobra [83]

9514 1404 393

Answer:

(x, y) = (-2.5, -4)

Step-by-step explanation:

Since you are given an expression for y, it is convenient to substitute that into the second equation.

2x -3(6x +11) = 7

2x -18x -33 = 7

-16x = 40 . . . . . . add 33

x = -40/16 = -2.5

Using the equation for y, we find ...

y = 6(-2.5) +11 = -15 +11 = -4

The solution is (x, y) = (-2.5, -4).

8 0

3 years ago

Help with this quistion

mariarad [96]

Answer:

Step-by-step explanation:

3 0

3 years ago

Solve for x <br> -1(-3x^2+2x+8)=1(0)

Igoryamba

Exact form: X = -4/3, 2
Decimal form: x= -1.3 repeat, 2

4 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