Answer:
no solution
Step-by-step explanation:
__________
The triangles PTQ and RTS are similar; so the ratios of corresponding sides are equal.
Let us denote the unknown length SQ as x.
We could make the following similarity relation;

Cross multiply to obtain;

Therefore;

Therefore, the answer is option A
its the first one because for example [-12] and 4
the absolute value of -12 is 12 so its greater
Answer:
The number of once is 9.1
The number of hundreds is 8.9
Step-by-step explanation:
Given as :
The total of digits having ones and hundreds = 900
The sum of digits = 18
Let The number of ones digit = O
And The number of hundreds digit = H
So, According to question
H + O = 18 .........1
100 × H + 1 × O = 900 ........2
Solving the equation
( 100 × H - H ) + ( O - O ) = 900 - 18
Or, 99 H + 0 = 882
Or , 99 H = 882
∴ H = 
I.e H = 8.9
Put the value of H in eq 1
So, O = 18 - H
I.e O = 18 - 8.9
∴ O = 9.1
So, number of once = 9.1
number of hundreds = 8.9
Hence The number of once is 9.1 and The number of hundreds is 8.9
Answer
Answer:
There area 14 one cent stamps
Step-by-step explanation:
Let
x -----> the number of 1 cent stamps
y -----> the number of 8 cent stamps
z -----> the number of 12 cent stamps
we know that
0.01x+0.08y+0.12z=1.78 -------> equation A
x=y+4 ------> y=x-4 -----> equation B
x=2z -----> z=0.5x ------> equation C
substitute equation B and equation C in equation A and solve for x
0.01x+0.08(x-4)+0.12(0.5x)=1.78
0.01x+0.08x-0.32+0.06x=1.78
0.15x=1.78 +0.32
0.15x=2.10
x=14
therefore
There area 14 one cent stamps