Dividing by a fraction is equivalent to multiply by its reciprocal, then:
Now, we need to express the quadratic polynomials using their roots, as follows:
where y1 and y2 are the roots.
Applying the quadratic formula to the first polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B%28-7%29%5E2-4%5Ccdot3%5Ccdot%28-6%29%7D%7D%7B2%5Ccdot3%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B121%7D%7D%7B6%7D%20%5C%5C%20y_1%3D%5Cfrac%7B7%2B11%7D%7B6%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B7-11%7D%7B6%7D%3D-%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the second polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot2%5Ccdot%28-3%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B25%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B5%7D%7B4%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-5%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the third polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B%28-3%29%5E2-4%5Ccdot2%5Ccdot%28-9%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B81%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B3%2B9%7D%7B4%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B3-9%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the fourth polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot1%5Ccdot%28-2%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B9%7D%7D%7B2%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B3%7D%7B2%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-3%7D%7B2%7D%3D-2%20%5Cend%7Bgathered%7D)
Substituting into the rational expression and simplifying:
Answer:
<B = 108 degrees
Step-by-step explanation:
<A = <B <u><em> (Alternate Angles are congruent)</em></u>
So,
6x+18 = x+93
=> 6x-x = 93-18
=> 5x = 75
<em>Dividing both sides by 5</em>
=> x = 15
Now,
<B = x+93
=> <B = 15+93
=> <B = 108 degrees
Answer:
The correct option is B.
Step-by-step explanation:
Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .
Draw a diagonal AC.
In triangle BCA and DAC,
AC\cong ACAC≅AC (Reflexive Property of Equality)
\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)
\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)
The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.
By ASA postulate,
\triangle BCA\cong \triangle DAC△BCA≅△DAC
Therefore option B is correct
Answer:
Perimeter of Δ ABC = 7 + 8 + 9 = 24 cm
Step-by-step explanation:
In triangle Δ XYZ ,
A is the mid point of XY
B is the midpoint of YZ
C is the mid point of XZ
AY = 7
BZ =8
XZ = 18
The mid - point theorem states that,
The segment formed by connecting two mid - points of a triangle is parallel to the third side and half as long
AY = 7 then BC = 7 cm
BZ = 8 then AC = 8 cm
XY = 18 then AB = 9 cm
Perimeter of Δ ABC = 7 + 8 + 9 = 24 cm
Answer:
3+5=5+3
Step-by-step explanation:
Because the commutative property means 'flip-flop'
Hope this helps! = )
