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Effectus [21]
3 years ago
5

Y=35+x how I do solve for x

Mathematics
1 answer:
Mazyrski [523]3 years ago
5 0

Try to put x on one side.

Do y       =       35+X

    -35            -35

Now you have Y-35=x

Hope this helped

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If the diameter of a circle is 5, what is its circumference?
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Multiply the diameter by PI which is 3.14 so that means the circumference would be 15.7.

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Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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Answer:

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From here, you can just solve for y, and it'll be in standard form. I would distribute 2/3 first.

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Then, add 1 to both sides.

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