<u><em>Answer: </em></u>
#9: 5
#10: -2
#11: -1.5
<em><u>Step-by-step explanation:</u></em>
<em><u>#9:</u></em> 3,8,13,18,23,(28),(33),(38),.. <em>Go up by 5 each time, so the common difference is </em><u><em>5</em></u><em>.</em>
<u><em>#10:</em></u> 11,9,7,5,3,(1),(-1),(-3),... <em>Go down </em><em>(-)</em><em> by 2 each time, so the common difference is </em><u><em>-2</em></u><em>.</em>
<u><em>#11:</em></u> 3, 1.5, 0, -1.5, -3, (-4.5), (-6), (-7.5),... <em>Go down </em><em>(-) </em><em>by 1.5 each time, so the common difference is </em><u><em>-1.5</em></u><em>.</em>
Answer:
![\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x--3%29%5E2%7D%7B49%7D%20-%5Cfrac%7B%28y-6%29%5E2%7D%7B32%7D%3D1)
Step-by-step explanation:
The standard equation of a horizontal hyperbola with center (h,k) is
![\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-h%29%5E2%7D%7Ba%5E2%7D-%5Cfrac%7B%28y-k%29%5E2%7D%7Bb%5E2%7D%3D1)
The given hyperbola has vertices at (–10, 6) and (4, 6).
The length of its major axis is
.
![\implies 2a=|14|](https://tex.z-dn.net/?f=%5Cimplies%202a%3D%7C14%7C)
![\implies 2a=14](https://tex.z-dn.net/?f=%5Cimplies%202a%3D14)
![\implies a=7](https://tex.z-dn.net/?f=%5Cimplies%20a%3D7)
The center is the midpoint of the vertices (–10, 6) and (4, 6).
The center is ![(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)](https://tex.z-dn.net/?f=%28%5Cfrac%7B-10%2B4%7D%7B2%7D%2C%5Cfrac%7B6%2B6%7D%7B2%7D%3D%28-3%2C6%29)
We need to use the relation
to find
.
The c-value is the distance from the center (-3,6) to one of the foci (6,6)
![c=|6--3|=9](https://tex.z-dn.net/?f=c%3D%7C6--3%7C%3D9)
![\implies 7^2+b^2=9^2](https://tex.z-dn.net/?f=%5Cimplies%207%5E2%2Bb%5E2%3D9%5E2)
![\implies b^2=9^2-7^2](https://tex.z-dn.net/?f=%5Cimplies%20b%5E2%3D9%5E2-7%5E2)
![\implies b^2=81-49](https://tex.z-dn.net/?f=%5Cimplies%20b%5E2%3D81-49)
![\implies b^2=32](https://tex.z-dn.net/?f=%5Cimplies%20b%5E2%3D32)
We substitute these values into the standard equation of the hyperbola to obtain:
![\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x--3%29%5E2%7D%7B7%5E2%7D%20-%20%5Cfrac%7B%28y-6%29%5E2%7D%7B32%7D%3D1)
![\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x%2B3%29%5E2%7D%7B49%7D%20-%5Cfrac%7B%28y-6%29%5E2%7D%7B32%7D%3D1)
Answer:
c.Right
Step-by-step explanation:
because of the 90 degrees it shows that it is 90 degrees
1,400 inches or 116.7 feet
![\bold{\huge{\purple{\underline{ Answers :- }}}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cpurple%7B%5Cunderline%7B%20Answers%20%3A-%20%7D%7D%7D%7D)
<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>
If I were one of the students in Barangay then I shall prepare the design of kite by using the known properties of kites in mathematics.
For example, Symmetrical about its main diagonals, Adjacent side equals, Having two pairs of congruent triangle etc.
<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u><u> </u></h3>
Design of kite assign to me
<u>Step </u><u>1</u><u> </u><u>:</u><u>-</u>
- I shall take one paper and cut it like that the adjacent sides of paper are equal
<u>Reason </u><u>:</u><u>-</u>
- <u>Adjacent </u><u>sides </u><u>of </u><u>kite </u><u>are </u><u>equal </u>
<u>Step </u><u>2</u><u> </u><u>:</u><u>-</u>
- I shall take two thin sticks and paste it on the paper but sticks should intersect each other at 90°
<u>Reason</u><u> </u><u>:</u><u>-</u>
- <u>Kite</u><u> </u><u>has </u><u>2</u><u> </u><u>diagonals </u><u>which </u><u>intersect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u><u> </u><u>.</u>
<u>Step </u><u>3</u><u> </u><u>:</u><u>-</u>
- <u>Make </u><u>a </u><u>hole </u><u>in </u><u>the </u><u>one </u><u>of </u><u>the </u><u>end </u><u>point </u><u>of </u><u>a </u><u>longest </u><u>sides</u><u>. </u>
<u>Observation </u><u>:</u><u>-</u>
- <u>The </u><u>kite </u><u>should </u><u>be </u><u>looked </u><u>like </u><u>that </u><u>it </u><u>having </u><u>two </u><u>pairs </u><u>of </u><u>congruent </u><u>triangle</u><u> </u><u>with </u><u>common </u><u>base. </u>
<h3><u>Answer </u><u>3</u><u> </u><u>:</u><u>-</u></h3>
- The adjacent sides of the kites are equal that is 4cm and 6cm
- The diagonals of the kite bisect each other at 90°
- As kite is symmetrical from main diagonals , so it has two opposite and equal Angles that is 127°
- The opposite angles at the end points of kite are congruent that is Angle D and Angle C
- AC is the bisector of AB and AB is the bisector of AC .
[ Note :- Kindly refer the above attachment ]
<h3><u>Answer </u><u>4</u><u> </u><u>:</u><u>-</u></h3>
All mathematical concepts used in making kite are as follows :-
- <u>Adjacent </u><u>sides </u><u>are </u><u>equal </u>
- <u>Diagonal </u><u>intersect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>
- <u>Having </u><u>two </u><u>pairs </u><u>of </u><u>congruent </u><u>triangle </u><u>with </u><u>common </u><u>base </u>
- <u>Symmetrical </u><u>about</u><u> </u><u>its </u><u>main </u><u>diagonal</u>
- <u>Opposite </u><u>angles </u><u>at </u><u>the </u><u>end </u><u>points </u><u>are </u><u>equal</u>