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Mamont248 [21]
3 years ago
5

On a rectangular soccer field, Sang is standing on the goal line 20 yards from the corner post. Jazmin is standing 99 yards from

the same corner post on the nearest adjacent side of the field. What is the distance from Sang to Jazmin?
A.119 yards
B.101 yards
C.10,201 yards
D.1,980 yards
Mathematics
1 answer:
Anna007 [38]3 years ago
3 0
Okay, so Sang is standing 20 yards away from one corner, and Jazmin is standing 99 yards away from the same corner. If this is a rectangle (I like visuals, so I'll use them to explain), then:
               
                    99ft
      A  -------------------------  B
         |                              |
20 ft  |                              |
         |                              |
    C   --------------------------  D

The question is asking you to solve for the diagonal line between points C and B. If you imagine a line there, you actually have the rectangle split into two triangles. So if you have triangle ABC, side CB would be the longest line, or the hypotenuse. That means you can use the Pythagorean Theorem to solve the problem.

A^2 + B^2 = C^2
99^2 + 20^2 = C^2
9,801 + 400 = C^2
10,201 = C^2

Now you solve for the square root of 10,201 to get C.

sqr (10,201) = C
C = 101 yards

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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
What is the area of the rhombus shown below? AC=17 BD=15 AB=11.2 Answers: A. 255 square units B. 190.4 square units C. 16 square
Lena [83]

Answer:

D. 127.5 square units

Step-by-step explanation:

AC=17

BD=15

AB=11.5

Diagonal (1&2) are given, base of the rhombus is also given.

Height is not given

Area of a rhombus given the diagonals

=1/2×d1×d2

Where,

d1=AC=17

d2=BD=15

Area of a rhombus=1/2×d1×d2

=1/2×17*15

=1/2×255

=127.5

D. 127.5 square units

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How to write 5.85 as a percent
Olenka [21]

The answer would be 585%

8 0
3 years ago
Read 2 more answers
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