Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
If you're just asking for how that'd look in algebraic form then the answer would be v+50/2
Answer:
3(x−4)(x+2)
Step-by-step explanation:
Answer:
303
Step-by-step explanation:
So the equation to find a term is An=a1+(n-1)d
An represents the value of the number (n)
and n is the selective that you want
d is the difference between the first and second term which is 23-18=5, and you can see that adding 5 to the previous term gives the following term.
a1 is the first term in the sequence
so knowing all that now you can go back and inset all variables into the equation
An=18+(58-1)5
An=303
Hope that helps :)
Please give brainliest
Let the four consecutive numbers be x,x+1,x+2,x+3.
Now the product of second and third number=(x+1)(x+2)=x²+3x+2=k
Product of first and fourth number=x(x+3)=x²+3x=k-2
A.T.Q
product of second and third number= product of first and fourth number+2
Hence proved
