Equation of the line:
y=mx+b
Ordinate at the origin (where the line cut the y-axis): b=1
B=(-5,5)=(xb,yb)→xb=-5, yb=5
C=(0,1)=(xc,yc)→xc=0, yc=1
m=(yc-yb)/(xc-xb)
m=(1-5)/(0-(-5))
m=(-4)/(0+5)
m=(-4)/(5)
m=-(4/5)
y=mx+b
y=-(4/5)x+1
Multiplying the equation by 5:
5[y=-(4/5)x+1]
5y=-4x+5
Adding 4x both sides of the equation:
5y+4x=-4x+5+4x
4x+5y=5
Answer: The equation for BC is: Third option 4x+5y=5
<span>Based on the problems I have recently done on FLVs with this though, is that if you connect the center of the arcs to the top intersection point, if the angle degrees, and opposite arcs are the same then it is an equilateral triangle.</span>
1. Angles ADC and CDB are supplementary, thus
m∠ADC+m∠CDB=180°.
Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.
2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then
m∠CDB=m∠CBD=65°.
The sum of the measures of interior angles of triangle is 180°, therefore,
m∠CDB+m∠CBD+m∠BCD=180° and
m∠BCD=180°-65°-65°=50°.
3. Triangle ABC is isosceles, with base BC. Then
m∠ABC=m∠ACB.
From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So
m∠ACB=65°.
4. Angles BCD and DCA together form angle ACB. This gives you
m∠ACB=m∠ACD+m∠BCD,
m∠ACD=65°-50°=15°.
Answer: 15°.
Answer:
your answer is b. 18.84ft
Step-by-step explanation:
Answer:
44 m/h
Step-by-step explanation:
speed/how fast = distance ÷ time
= 308 miles ÷ 7 hours
= 44m/h
