Answer:
18) Area= 5*5/2=25/2=12.5 unit ^2
19) Area=AB^2V3/4=8a^2*V3/4=2V3a^2
Step-by-step explanation:
18. A(-3,0)
B(1,-3)
C(4,1)
AB=V(-3-1)^2+(0+3)^2=V16+9=V25=5
AC=V(-3-4)^2+(0-1)^2=V49+1=V50=5V2
BC=V(1-4)^2+(-3-1)^2=V9+16=V25=5
so AB=BC=5
and AC^2=AB^2+BC^2
so trg ABC is an isosceles right angle triangle (<B=90)
Area= 5*5/2=25/2=12.5 unit ^2
19. A(a,a)
B(-a,-a)
C(-V3a, V3a)
AB=V(a+a)^2+(a+a)^2=V4a^2+4a^2=V8a^2
AC=V(a+V3a)^2+(a-V3)^2=Va^2+2a^2V3+3a^2+a^2-2a^2V3+3a^2=V8a^2
BC=V(-a+V3a)^2+(-a-V3a)^2=V8a^2
so AB=AC=BC
Area=AB^2V3/4=8a^2*V3/4=2V3a^2
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D
The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3
Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
Answer:
On a coordinate plane, a triangle has points (negative 4, 3), (negative 4, negative 2), (1, negative 2).
Step-by-step explanation:
The points (-4,3), (1,2) and (-4, -2) would form a right triangle when graphed and connected by lines.
(-4,3), (1,2) and (1,3) would also work as well
Answer:idk im
Step-by-step explanation:
