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2 years ago

This graph shows the solution to which inequality? Apex

Mathematics

2 answers:

Veronika [31]2 years ago

5 0

The answer to this question is "D". To understand where to shade, just plug in (0,0) to the linear equation. when you plug in (0,0) it makes your last option true

just olya [345]2 years ago

4 0

Answer:

Option D. y > $\frac{4}{3}$ x - 2

Step-by-step explanation:

Before analyzing the graph of an inequality we should always keep in our mind.

(1) Dotted lines shows the sign of inequality ( greater than or less than ) and solid line shows the sign of equality.

(2) Shaded area above the line is greater than and below the line is less than.

Now in the given figure dotted line with shaded area above the line will be represented by

y > $\frac{4}{3}$ x - 2

Option D. is the answer.

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Explain what is 90% of 180

Yakvenalex [24]

Answer:

The answer is 162.

Step-by-step explanation:

To find the percentage of a number:

Turn the percentage into a decimal. 90% to .90

Multiply the decimal by the number you want to find the percentage of. .90 x 180

The answer is the product.

4 0

3 years ago

Which pair of angles are an example of alternate exterior angles?

9966 [12]

Answer:

c. is the answer becuase since it is exterior they are talking about the outside of the parellel lines.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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Eva8 [605]

Answer:

initial value=2

rate of change=1

Step-by-step explanation:

rate of change=(3-2)/(1-0)=1

7 0

3 years ago

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0

2 years ago

Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8

Dima020 [189]

Let $f(x,y,z)=x+y+z-8e^{xyz}$ . The tangent plane to the surface at (0, 0, 8) is

$\nabla f(0,0,8)\cdot(x,y,z-8)=0$

The gradient is

$\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)$

so the tangent plane's equation is

$(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8$

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

$(1,1,1)t+(0,0,8)=(t,t,t+8)$

or $x(t)=t$ , $y(t)=t$ , and $z(t)=t+8$ .

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)

4 0

2 years ago