Answer:
Option (2).
Step-by-step explanation:
It is given in the question,
ΔLMN is a right triangle with base LM = 3a units
Hypotenuse MN = 5a
By applying Pythagoras theorem in ΔLMN,
MN² = LM² + NM²
(5a)² = (3a)² + MN²
25a² - 9a² = MN²
MN = √16a²
MN = 4a
Therefore, vertices of the triangle will be L(0, 0), M(3a, 0) and N(0, 4a).
Option (2) will be the answer.
Triangles congruent by ASA have two pairs of congruent sides and an included congruent angle.
The graph indicates that sides TV, HG, and AB are congruent, and that sides TU, FG, and BC are congruent. It also indicates that angles U, F, and C are congruent, and that angles G and B are congruent. Notice that angle U of triangle TUV is not an included angle; this eliminates triangle TUV as it can't be congruent to another triangle by ASA with the information provided.
That leaves triangles FGH and ABC. Evidently, angles G and B are included angles, so these triangles are congruent by ASA.
Answer:
b. ΔHGF and ΔABC
Answer:
5
Step-by-step explanation:
a polynomial has one quadratic factor and 3 linear factors. One of the linear factors has multiplicity two. What is the degree of the polynomial
A polynomial with one quadratic obtains the forms ( ax² +bx +c ) with 3 linear factors.
Suppose the three linear fractions are :
(x- P) (x-Q) (x- R)
∴
The polynomial = ( ax² +bx +c )(x- P) (x-Q) (x- R)
By factorization, the highest degree of the polynomial = 5
Answer:
I beleive that the answer is (1,2)
Step-by-step explanation:
-8x - 60 = -164
Add 60 to both sides, because whatever you do on one side, you have to do it to the other side
-8x = -104
Divide -8 to both sides
x = 13
