Add one third to 5 times d
1 answer:
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Answer:
(Angle DBE) ∠DBE = 50°
Step-by-step explanation:
It is crucial for us to be able to interpret plane geometry related question in order to solve them efficiently.
From the question given; we have analysed and come up with a diagrammatic expression and mathematical solution that clearly explains the question .
Given that
BC and DF are parallel lines.
B is a point on AD
B is a point on AD
BD=BE
work out angle DBE and give a reason.
The diagram can be seen in the attached image below.
SInce ∠BD = ∠DE
Then triangle BDE is an isosceles triangle. In an isosceles triangle ; two sides are equal in length.
The sum of angles in an isosceles triangle = 180° (sum of angles in a triangle)
So;
∠BDE + ∠BED + ∠DBE = 180° (sum of angles in a triangle)
From the diagram; we will see that ∠ABC = ∠BDE (corresponding angle)
Since ; ∠ABC is not given and which is needed to solve this question.
Let's just assume that ∠ABC is 65° , the main thing is to be able to interpret and understand the concept of the question.
Now;
Since
∠ABC = ∠BDE
∠ABC = 65°
∠BDE = 65°
Again;
∠BDE + ∠BED + ∠DBE = 180° (sum of angles in a triangle)
∠BDE will be aso equal to ∠BED ; this is because since the length of the opposite sides of the isosceles triangle are equal, their angles will also be equal.
Therefore;
65° + 65° + ∠DBE = 180° (sum of angles in a triangle)
130 ° + ∠DBE = 180° (sum of angles in a triangle)
∠DBE = 180° - 130°
∠DBE = 50°
Answer:
11.4
20/100 = x/57
Step-by-step explanation:
It is perhaps simplest just to translate the given English sentence into a mathematical sentence.
... 20% ... ... of ... 57 ... is ... what number
... 20/100 ... × ... 57 .... = ... what number . . . . . . . . translation to math sentence
... 11.4 = what number . . . . . . . . carry out the multiplication
_____
If you're asked to write this as a proportion, you can write it like this:
You can solve this proportion by multiplying by 57 (the denominator under "what number"). Doing so gives you the same arithmetic sentence as shown above.