If DEF=GJH, find each missing measure
a) DF =
d) F =
b) JH =
|||
e) m&G =
c) GJ =
f) m =
I need help it’s due in 20 minutes:(
Answers
Congruent triangles have equal corresponding sides and angles.
The measure of the side lengths are: [tex]\mathbf{DF = 16m}[/tex], [tex]\mathbf{JH = 14m}[/tex] and [tex]\mathbf{GJ = 19m}[/tex]The measure of the angles are: [tex]\mathbf{\angle G = 34^o}[/tex], [tex]\mathbf{\angle J = 41^o}[/tex] and [tex]\mathbf{\angle F = 105^o}[/tex]The given parameters are:
[tex]\mathbf{\triangle D EF \cong \triangle GJH}[/tex]
[tex]\mathbf{EF = 14m}[/tex]
[tex]\mathbf{D E = 19m}[/tex]
[tex]\mathbf{GH = 16m}[/tex]
[tex]\mathbf{\angle D = 34^o}[/tex]
[tex]\mathbf{\angle H = 105^o}[/tex]
[tex]\mathbf{\triangle D EF \cong \triangle GJH}[/tex] means that triangles DEF and GHJ are congruent triangles.
So, we have:
[tex]\mathbf{DF = GH}[/tex]
This gives
[tex]\mathbf{DF = 16m}[/tex]
Also, we have:
[tex]\mathbf{JH = EF}[/tex]
This gives
[tex]\mathbf{JH = 14m}[/tex]
Also, we have:
[tex]\mathbf{GJ = DE}[/tex]
This gives
[tex]\mathbf{GJ = 19m}[/tex]
The following angles are also congruent
[tex]\mathbf{\angle F = \angle H}[/tex]
This gives
[tex]\mathbf{\angle F = 105^o}[/tex]
Also, we have:
[tex]\mathbf{\angle G = \angle D}[/tex]
This gives
[tex]\mathbf{\angle G = 34^o}[/tex]
The measure of angle J is then calculated using
[tex]\mathbf{\angle J + \angle H + \angle G = 180^o}[/tex]
This gives
[tex]\mathbf{\angle J + 105 + 34= 180^o}[/tex]
[tex]\mathbf{\angle J + 139= 180^o}[/tex]
Subtract 139 from both sides
[tex]\mathbf{\angle J = 41^o}[/tex]
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Step-by-step explanation: hope it helps
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The coefficient of x is 15.
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What is an expression?An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.
Example: 2 + 3x + 4y = 7 is an expression.
We have,
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[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\textsf{Given expression;}\\[/tex]
[tex] \rm{23 = \frac{7}{9}(6x - 36) + 9 } \\ [/tex]
[tex] \rm{\longmapsto} \: 23 = \frac{(42x - 256)}{9} + 9 \\ [/tex]
[tex] \rm{\longmapsto} \: 23 = \frac{(42x - 256 + 81)}{9} \\ [/tex]
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[tex] \rm{\longmapsto} \: \frac{23}{1} = \frac{(42x - 175)}{9} \\ [/tex]
[tex]\textsf{By doing cross multiplication, we get}\\[/tex]
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----------------
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----------------
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