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3 years ago

The system of equations is graphed on the coordinate plane. y=−1/2x−1y=1/4x−4

Mathematics

2 answers:

nikdorinn [45]3 years ago

4 0

(4,-3) i took the test i'm 100% sure
i hope i helped ;)

Yuki888 [10]3 years ago

4 0

Im just completed the test the correct answer is 4,-3
Please mark as brainiest :)

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Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0) is a saddle point

C(3,0) is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is

$f(x,y)=x^3+y^3-3x^2-3y^2-9x$

The first partial derivative with respect to x is

$f_x=3x^2-6x-9$

The first partial derivative with respect to y is

$f_y=3y^2-6y$

We now set each equation to zero to obtain the system of equations;

$3x^2-6x-9=0$

$3y^2-6y=0$

Solving the two equations simultaneously, gives;

$x=-1,x=3$ and $y=0,y=2$

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

$D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$

But, we would have to calculate the second partial derivatives first.

$f_{xx}=6x-6$

$f_{yy}=6y-6$

$f_{xy}=0$

$\Rightarrow D=(6x-6)(6y-6)-0^2$

$\Rightarrow D=(6x-6)(6y-6)$

At A(-1,0),

$D=(6(-1)-6)(6(0)-6)=72\:>\:0$ and $f_{xx}=6(-1)-6=-18\:$

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

$D=(6(-1)-6)(6(2)-6)=-72\:$

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

$D=(6(3)-6)(6(0)-6)=-72\:$

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

$D=(6(3)-6)(6(2)-6)=72\:>\:0$ and $f_{xx}=6(3)-6=12\:>\:0$

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0

3 years ago

I need all 4 help pls

slavikrds [6]

Answer:

I dont know the answer wait ima get it right now

Step-by-step explanation

i do not know the answer

3 0

2 years ago

Read 2 more answers

Help me on this question! Jake can carry 6 1/4 pounds of wood in from the barn.His father can carry 1 5/7 times as much as jake.

ioda

$\text{ Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }$

Solution:

From given question,

Number of pounds Jake carry = $6\frac{1}{4} \text{ pounds }$

Number of pounds his father carry is $1\frac{5}{7}$ times as much as jake

To find: Number of pounds Jake father can carry

Convert the mixed fractions to improper fractions

Multiply the whole number part by the fraction's denominator.

Add that to the numerator.

Then write the result on top of the denominator

$\rightarrow 6\frac{1}{4} = \frac{4 \times 6+1}{4} = \frac{25}{4}\\\\\rightarrow 1\frac{5}{7} = \frac{7 \times 1+5}{7} = \frac{12}{7}$

Then according to question,

$\text{Pounds father carry } = \frac{12}{7} \text{ times the pounds jake carry }\\\\\text{Pounds father carry } = \frac{12}{7} \times \frac{25}{4}\\\\\text{Pounds father carry } = \frac{75}{7}\\\\\text{In terms of mixed fractions, }\\\\\text{Pounds father carry } = \frac{75}{7} = 10\frac{5}{7}$

$\text{ Thus, Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }$

3 0

3 years ago

Someone help need this done right now I’ll mark u brainliest

Ad libitum [116K]

Answer:

89 meters

Step-by-step explanation:

We know it's 9 + "the height of the triangle".

Fortunately, the height of the triangle (let's call it y) can be computed from:

y^2 + 18^2 = 82^2 (Pythagoras theorem).

y^2 = 82^2 - 18^2 = 6724 - 324 = 6400

y = 80 (or y = -80 but it doesn't make sense in geometry).

so x = y + 9 = 89;

3 0

3 years ago

It has been estimated that a student's final grade in a course decreases by 5 points for every 2 days absent from class. If a st

OverLord2011 [107]

Answer:

the final grade will decrease by 10 points

3 0

3 years ago