Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
Multiply the bracket by 5
I used PEMDAS
P= parenthesis
E= exponents
M=multiplication
D= division
A= addition
S= subtraction
7r+2= 5(r-4)
7r+2= 5r-20
Move 5r to the left hand side . Positive 5r changes to negative 5r
7r-5r+2= 5r-5r-20
2r+2=- 20
2r+2-2= -20-2
Move positive 2 to the right hand side. Changes to negative -2
2r+2-2= -20-2
2r= -22
Divide by 2 for 2r and -22
2r/2= -22/2
r= -11
Answer is r= -11
Answer:
Step-by-step explanation:
Given
The attached graph
Required
Determine the line equation
First, list out two points from the graph
Next, calculate the slope (m)
The equation in slope intercept form is:
This gives:
Open bracket
Answer:
12 = 22 × 3
Step-by-step explanation:
Divide <span>150:100 to find out what the density in g/cm3 is. So 150:100 = 1.5 g/cm3</span>
