DE = AB, EF = BC and AC = DF, hence triangle ABC is congruent to triangle DEF.
<h3>
Congruent shape</h3>
Two shapes are said to be congruent if they have the same shape, all their corresponding angles and sides are congruent to one another.
Given that DE = AB and BC = EF.
In right triangle DEF, using Pythagoras:
DF² = DE² + EF²
Also, In right triangle ABC, using Pythagoras:
AC² = AB² + BC²
But DE = AB and EF = BC, hence:
AC² = DE² + EF²
AC² = DF²
Taking square root of both sides, hence:
AC = DF
Since DE = AB, EF = BC and AC = DF, hence triangle ABC is congruent to triangle DEF.
Find out more on Congruent shape at: brainly.com/question/11329400
7√27 + 5√48
7 × 3√3 + 5√48
7 × 3√3 + 5 × 4√3
21√3 + 5 × 4√3
21√3 + 20√3
41√3 or 71.014
hope this helps, God bless!
Step-by-step explanation:
I don't know if it's correct
Step-by-step explanation:
Start by multiplying both sides by cosα:
1 + sinα + (cos²α)/(1+sinα) = 2
sinα + (cos²α)/(1+sinα) = 1
Now multiply both sides by 1+sinα:
sinα + sin²α + cos²α = 1 + sinα
sin²α + cos²α = 1 Q.E.D.
-8n+7=31
-8n+7-7=31-7
-8n=31-7
-8n=24
-8n÷(-8)=24÷(-8)
n=24÷(-8)
n=-24÷8
n=-3
