DOMAIN:
- Set of allowable inputs (set of 1st elements of the ordered pairs in relation)
- X point
- AKA the input
DEFINE INPUT:
A number that'll be used in a function rule to determine the value of the output
RANGE:
- Set of possible outputs (set of 2nd elements of the ordered pairs in relation)
- Y point
- AKA the output
DEFINE OUTPUT:
Result of applying a function rule to the value of ani input
ORDERED PAIRS:
(-1, 7): -1 is the domain & 7 is the range
(3, 5): 3 is the domain & 5 is the range
(4, 9): 4 is the domain & 9 is the range
(-6, -10): -6 is the domain & -10 is the range
(-2, 8): -2 is the domain & 8 is the range
Hope this helps you!!! :)
Answer:
see explanation
Step-by-step explanation:
Inequalities of the type | x | > a always have solutions of the form
x < - a or x > a, thus
2x + 1 < - 5 OR 2x + 1 > 5 ( subtract 1 from both sides of both inequalities
2x < - 6 OR 2x > 4 ( divide both sides of both inequalities by 2 )
x < - 3 OR x > 2
Answer:
x1=-4/5, x2=18/5 and x3=7/5
Step-by-step explanation:
![\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C3%261%262%264%5C%5C2%262%261%267%5Cend%7Barray%7D%5Cright%5D)
you can do linear combination between the rows:
2nd row=R2-3R1 and 3th row=R3-2R1
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%262%267%2617%5Cend%7Barray%7D%5Cright%5D)
3th row=(3R2-R3)/15
![\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%26-3%26-5%5C%5C0%261%2611%2619%5C%5C0%260%261%26%5Cfrac%7B7%7D%7B5%7D%20%5Cend%7Barray%7D%5Cright%5D)
1st row=R1+3R3 and R2-11R3
![\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-4%2F5%20%5C%5C0%261%260%2618%2F5%20%5C%5C0%260%261%267%2F5%5Cend%7Barray%7D%5Cright%5D)
x1=-4/5, x2=18/5 and x3=7/5
<u>Explanation</u><u>:</u>
Consider ABCD is a rhombus
We know that
All sides are equal in rhombus i.e,
⇛AB=BC=CD=DA
and AC and BD are digonals
Given that
Diagonal and the side of the rhombus are equal.
⇛AB = BC = CD = DA = AC
Diagonal AC divides the rhombus into two triangles .
They are ∆ BAC and ∆ DAC
In triangle BAC
BA=BC=AC,(Given)
⇛∠ BAC=∠ABC= ∠ACB =60°→→→Eqn(i)
Similarly in ∆DAC ,
DA=DC=AC
⇛∠DAC=∠ACD=∠ADC=60°→→→Eqn(ii)
From eqn(i) and eqn(ii)
∠A=∠BAC+∠DAC=60°+60°=120°
and
∠B= ∠ABC = 60°.
and
∠C=∠ACB+∠ACD=60°+60°=120°
and
∠D =∠ADC=60°
∴ ∠A = 120° , ∠B = 60° ,∠C = 120° & ∠D = 60°
<u>Answer:</u><u>-</u>The measures of the all angles in the rhombus are 120° , 60° ,120° and 60°.
Note: [Figure refers in the attached file.
