m∠HDG = 28°, m∠EFG = 50°, m∠DEG = 67°, m∠DGE = 65°
Solution:
Triangle sum property:
Sum of the angles of the triangle = 180°
In ΔDHG,
m∠HDG + 120° + 32° = 180°
m∠HDG + 152° = 180°
m∠HDG = 180° – 152°
m∠HDG = 28°
In ΔGEF,
m∠EFG + 17° + 113° = 180°
m∠EFG + 130° = 180°
m∠EFG = 180° – 130°
m∠EFG = 50°
Sum of the adjacent angles in a straight line is 180°
m∠DEG + m∠DEF = 180°
m∠DEG + 113° = 180°
m∠DEG = 180° – 113°
m∠DEG = 67°
In ΔDGE,
m∠DGE + 48° + 67° = 180°
m∠DGE + 115° = 180°
m∠DGE = 180° – 115°
m∠DGE = 65°
Hence m∠HDG = 28°, m∠EFG = 50°, m∠DEG = 67°, m∠DGE = 65°.
Answer:12 9/10
Step-by-step explanation:
Answer:
it would stay the same
Step-by-step explanation:
For functions don’t repeat the x (x,y)
It is B
Answer:
8x² +36x -27=0
Step-by-step explanation:
2x²= 6x +3
Let's rewrite the equation into the form of ax²+bx+c= 0.
2x² -6x -3=0
Thus, a= 2
b= -6
c= -3
Sum of roots= 
Since your roots are p and q, sum of roots= p +q
p +q= 
p +q= 6 ÷2
p +q= 3
Product of roots= 
pq= 
<u>Quadratic equations</u><u>:</u>
x² -(sum of roots)x +(product of roots)= 0
Thus, we have to find the sum and the product of the new roots, p²q and pq².
p +q= 3
pq= -3/2
Product of new roots
= (p²q)(pq²)
= p³q³
= (pq)³

sum of new roots
= p²q +pq²
= pq(p +q)
= (-3/2)(3)
= -9/2
Thus, the quadratic equation with roots p²q and pq² is
x² -(-9/2)x -27/8 = 0

Multiply by 8 throughout:
