The diagonal of a rhombus divides it into two congruent isosceles triangles.
So ∠CBD ≅ ∠CDB
∠CBD + ∠CDB + 68 = 180
2∠CBD = 180 - 68 = 112
∠CBD = 56
We also have
∠BDE + ∠E + ∠DBE = 180
∠DBE = 180 - 73 - 36 = 71
∠EBC = ∠EBD - ∠CBD = 71 - 56 = 15
Answer: ∠EBC = 15 degrees
The product of 3p and q-3 would be put in to equation form like this: 3p(q-3)
To find your answer. You have to distribute the 3p to by individually multiplying it by q and -3
It should now look like this:
So your answer is:
Since you have the answers given, just track their walking.
After one hour, one guy has covered 3mi and the other 4mi, which is in total 7, not 14
After 2 hours, one guy has covered 6mi and the other 8mi, which is 14mi
The algebraic way would be:
3x +4x = 14 *x being the hour
7x = 14
x =2
I thought of it as answering a regular equation and imagine it's equal signs but when your done solving you with have like ex x<5 and where ever the butt of the arrow is pointing is where the lines goes if there's ano equal sign below it then it is includive
X=2h, y=3k
Substitute these values into equations.
y+2x = 4 ------> 3k+2*2h=4 -----> 3k +4h =4
2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1
We have a system of equations now.
3k +4h =4 ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1
2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0
4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0
(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0
Denominator cannot be = 0
4h(2-2h)≠0
Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4 or h=-1
Check which
4h(2-2h)≠0
1) h= 3/4 , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.
h=3/4, then 3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then 3k = 4-4*(-1) =8 , 3k=8, k=8/3
So,
if h=3/4, then k=1/3,
and if h=-1, then k=8/3 .
