Answer:
2x + 3 = 9
Step-by-step explanation:
Three more than twice a number is nine
Twice a number would be 2x
where x is the number
Three more would be + 3
And is means equal sign or “=“
Yes he is correct because 1/3 can be shown on a number line
Answer:
Sure why not.
Step-by-step explanation:
Answer:
x= -3 and y= 0
Step-by-step explanation:
5x+2y=-15
<u>2x-2y=-6 </u>
<u>7x =-21</u>
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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C2%26-2%5C%2F%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let
= A = X and = 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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-2%5C%5C-2%265%5C%2F%5Cend%7Barray%7D%5Cright%5D)
║A║= ( 5*-2- 2*2)= -10-4= -14≠0
X= A⁻¹ B
=- 1/14
=- 1/14 ![\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2A-15%26%2B%20-2%2A-6%5C%5C-2%2A-15%26%2B%205%2A-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2030%26%2B12%5C%5C30%26%2B-30%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc}42\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D42%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-42%2F14%5C%5C0%2F-14%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
From here x= -3 and y= 0
Solution Set = [(-3,0)]
Answer:
A) AAS; B) LA; C) ASA
Step-by-step explanation:
AAS is the Angle-Angle-Side congruence statement. It says that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of a second triangle, then the triangles are congruent. In these triangles, ∠E≅∠K, ∠F≅∠L, and DE≅JK. These are two angles and a non-included side; this is AAS.
LA is the leg-acute theorem. It states that if a leg and acute angle of one triangle is congruent to the corresponding leg and acute angle of another triangle, then the triangles are congruent.
The leg we have congruent from each triangle is DE and JK. We also have ∠E≅∠K and ∠F≅∠L, both pairs of which are acute. This is the LA theorem.
ASA is the Angle-Side-Angle congruence statement. It says that if two angles and an included side of one triangle are congruent to the corresponding two angles and included side of another triangle, then the triangles are congruent.
We have that ∠D≅∠J, DE≅JK and ∠E≅∠K. This gives us two angles and an included side, or ASA.
