Lisa had a starting balance in her checking account of $956.02. She

Which of the following inequalities matches the graph?

zaharov [31]

Answer: the inequality that matches de graph is x ≤ 5.

Explanation:

1) The vertical line through the point (5,0) is represented by x = 5.

2) The area to the left of the vertical line x = 5 contents all the points whose x-coordinate is less than five.

Please, find attached the figure with the graph described in the question.

3) A solid line is used to indicate that the points on the line are included in the solution region (a dashed line is used to indicate that the points on the line are excluded)

Therefore, the graph includes the points on the line x = 5 and the points with x-coordinate less than 5.

4) Hence, you have to use an inequality with the symbol ≤, which means less than or equal to.

x ≤ 5 means less than or equal to 5, which is the inequality searched.

4 0

3 years ago

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A veterinarian predicted that the weight of a newborn puppy would be 1.5 pounds. When the dog gave birth, the newborn weighed 2

shtirl [24]

Answer:

i wanna be christen

Step-by-step explanation:

howww

5 0

3 years ago

What is the value of h? h = 20 h = 35 h = 55 h = 70

Yuri [45]

Answer:55

Step-by-step explanation:

8 0

2 years ago

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

Find the length of the third side. If necessary, round to the nearest tenth.

schepotkina [342]

Answer:

4.5

Step-by-step explanation:

To find the hypotenuse of a right triangle, we use Pythagorean's Theorem which states that $a^{2}+b^{2} =c^{2}$ . Plugging in the values of the two legs given, we get $2^{2} +4^{2} =c^{2} =20$ . We can take the square root of each side and get that $c=\sqrt{20} =4.5$ .

4 0

2 years ago

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