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kondor19780726 [428]
4 years ago
13

Maurice and Johanna have appreciated the help you have provided them and their company Pythgo-grass. They have decided to let yo

u consult on a big project.
1.A triangular section of a lawn will be converted to river rock instead of grass. Maurice insists that the only way to find a missing side length is to use the Law of Cosines. Johanna exclaims that only the Law of Sines will be useful. Describe a scenario where Maurice is correct, a scenario where Johanna is correct, and a scenario where both laws are able to be used. Use complete sentences and example measurements when necessary.

2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.

3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.

5.Two pillars have been delivered for the support of a shade structure in the backyard. They are both ten feet tall and the cross-sections​ of each pillar have the same area. Explain how you know these pillars have the same volume without knowing whether the pillars are the same shape.
Mathematics
1 answer:
Colt1911 [192]4 years ago
5 0
<span><span>1.A triangular section of a lawn will be converted to river rock instead of grass. Maurice insists that the only way to find a missing side length is to use the Law of Cosines. Johanna exclaims that only the Law of Sines will be useful. Describe a scenario where Maurice is correct, a scenario where Johanna is correct, and a scenario where both laws are able to be used. Use complete sentences and example measurements when necessary.
</span>
The Law of Cosines is always preferable when there's a choice.  There will be two triangle angles (between 0 and 180 degrees) that share the same sine (supplementary angles) but the value of the cosine uniquely determines a triangle angle.

To find a missing side, we use the Law of Cosines when we know two sides and their included angle.   We use the Law of Sines when we know another side and all the triangle angles.  (We only need to know two of three to know all three, because they add to 180.  There are only two degrees of freedom, to answer a different question I just did.

<span>2.An archway will be constructed over a walkway. A piece of wood will need to be curved to match a parabola. Explain to Maurice how to find the equation of the parabola given the focal point and the directrix.
</span>
We'll use the standard parabola, oriented in the usual way.  In that case the directrix is a line y=k and the focus is a point (p,q).

The points (x,y) on the parabola are equidistant from the line to the point.  Since the distances are equal so are the squared distances.

The squared distance from (x,y) to the line y=k is </span>(y-k)^2
<span>
The squared distance from (x,y) to (p,q) is </span>(x-p)^2+(y-q)^2.<span>
These are equal in a parabola:

</span>
(y-k)^2 =(x-p)^2+(y-q)^2<span>

</span>y^2-2ky + k^2 =(x-p)^2+y^2-2qy + q^2

y^2-2ky + k^2 =(x-p)^2 + y^2 - 2qy+ q^2

2(q-k)y =(x-p)^2+ q^2-k^2

y = \dfrac{1}{2(q-k)} ( (x-p)^2+ q^2-k^2)

Gotta go; more later if I can.

<span>3.There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane.

4.A pipe needs to run from a water main, tangent to a circular fish pond. On a coordinate plane, construct the circular fishpond, the point to represent the location of the water main connection, and all other pieces needed to construct the tangent pipe. Submit your graph. You may do this by hand, using a compass and straight edge, or by using a graphing software program.

5.Two pillars have been delivered for the support of a shade structure in the backyard. They are both ten feet tall and the cross-sections​ of each pillar have the same area. Explain how you know these pillars have the same volume without knowing whether the pillars are the same shape.</span>
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The point A(-3, -2) is translated using T: (x,y) → (x + 5, y - 3). What is the distance from A to A'?
Degger [83]

Answer:

d=5.39 units

Step-by-step explanation:

Given A (-3, -2)  

T= (x+5, y-3)

T= (-3+5, -2-3)

T=(2,-5)

Applying the translation on;

A( -3,-2)

A' (-3+2, -2+ -5)

A'= (-1, -7)

Distance

d= \sqrt { X2-X1)^2 + (Y2-Y1)^2

A = (-3,-2)     and  A'(-1, -7)

d=√ (-1--3)² + (-7--2)²

d=√ (2)²+(-5)²

d=√4+25

d=√29

d=5.39 units

4 0
3 years ago
7 3/9 - 2 4/9 =<br>11 3/4 - 5 1/4 =​
jeka57 [31]

1.)

7 \frac{3}{9}  - 2 \frac{4}{9}  \\  =  \frac{66}{9}  -  \frac{22}{9}  \\  =  \frac{44}{9}  \\  = 4 \frac{8}{9}

2.)

11 \frac{3}{4}  - 5 \frac{1}{4}  \\  =  \frac{47}{4}  -  \frac{21}{4}  \\  =  \frac{26}{4}  \\  =  \frac{13}{2}  \\  = 6 \frac{1}{2}

<em>Hope it helps and is useful :)</em>

6 0
3 years ago
Solve for x????????????????????
12345 [234]

\huge \tt \color{pink}{A}\color{blue}{n}\color{red}{s}\color{green}{w}\color{grey}{e}\color{purple}{r }

\large\underline{  \boxed{ \sf{✰\: Concept }}}

  • ➣ Given two paris are linear pairs
  • ➣ We know that sum of linear pairs is equal to 180⁰
  • ➣ Linear :- Having the form of a line; straight or roughly straight; following a direct course.

\large\underline{  \boxed{ \sf{✛\: Assumptions\:✛ }}}

  • ➣ let given two angles be p and q that is
  • ➣ angle p is 3x-4
  • ➣ angle q is 3x-8
  • ➣ we have to find "x" in the given expressions

\large\underline{  \boxed{ \sf{✰\: let's\:solve }}}

\rm\angle \: p +  \angle \: q = 180 \degree \\

★ Substituting values

\rm { \pink➾ \:( 3x  - 4) + (3x - 8) = 180} \\ \rm { \pink➾}arranging \: and \: solving \: like \: terms \\ \rm { \pink➾ \: 3x + 3x - 4 - 8 = 180} \\ \rm { \pink➾ \: 6x - 12 = 180} \\ \rm  { \pink➾6x = 180  + 12}

\qquad \qquad\rm { \pink➾6x  = 192} \\ \qquad\qquad\rm { \pink➾ \: x =  \frac{ \cancel{192}}{ \cancel6} } \\ \qquad\qquad\rm { \pink➾x = 32}

★ Hence the value

\boxed{↪\underline{  \boxed{ \rm{\: x = 32\green✓}}}↩}

\rule{70mm}{2.9pt}

Hope it helps !

5 0
2 years ago
The ratio of books to magazines in the classroom is 7:3. If there are 28 books in the
ss7ja [257]

Answer:

Step-by-step explanation:

7 books.................................3 magazines

14 books.................................6 magazines

21 books................................9 magazines

28 books..............................12 magazines

of course we ca do it

7/3=28/x

x=(3*28)/7=12

6 0
3 years ago
suppose the probality that it will rain tomorrow is 0.59. what is the probality that it will not rain tomorrow​
kykrilka [37]

Answer: 0.41

Step-by-step explanation: 0.59 is 59%. To find the percentage of chance that it will not rain, you subtract 1 (which is 100%) minus 0.59. You get an answer of 0.41 (or 41% if you prefer.)

7 0
3 years ago
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