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Tpy6a [65]
2 years ago
5

In the diagram, the straight line ABC is parallel to EFG and DB is parallel to FC. Given that ABD=38 degrees and DFE=62 degrees.

Find angle BDF and CFG
Mathematics
1 answer:
Elanso [62]2 years ago
4 0

Based on the calculations, the measure of angle BDF and CFG are 100° and 38° respectively.

<h3>The condition for two parallel lines.</h3>

In Geometry, two (2) straight lines are considered to be parallel if their slopes are the same (equal) and they have different y-intercepts. This ultimately implies that, two (2) straight lines are parallel under the following conditions:

m₁ = m₂

<u>Note:</u> m is the slope.

<h3>What is the alternate interior angles theorem?</h3>

The alternate interior angles theorem states that when two (2) parallel lines are cut through by a transversal, the alternate interior angles that are formed are congruent.

Based on the alternate interior angles theorem, we can infer and logically deduce the following properties from the diagram (see attachment):

  • <ABD = <BDH = 38°
  • <DFE = <HDF = 62°

For angle BDF, we have:

<BDF = <BDH + <HDF

<BDF = 38° + 62°

<BDF = 100°.

Since angles BDF and DFC are linear pair, they are supplementary angles. Thus, we have:

∠BDF + <DFC = 180°

<DFC = 180 - ∠BDF

<DFC = 180 - 100

<DFC = 80°.

For angle CFG, we have:

∠DFE + <DFC + <CFG= 180°

<CFG = 180° - ∠DFE - <DFC

<CFG = 180° - 62° - 80°

<CFG = 38°.

Read more on parallel lines here: brainly.com/question/3851016

#SPJ1

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Answer:

Sope = mx+b

Step-by-step explanation:

Slope=-5/4x

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3 years ago
The length of a rectangular garden is 6 more than its width. If the perimeter is 32 feet, what is the width and the area of the
PolarNik [594]

Answer:

{\fbox{ \sf{Width \: of \: a \: rectangular \: garden \:  = 5 \: feet}}}

\boxed{ \sf{ \: Area \: of \: a \: rectangular \: garden = 55 {ft}^{2}}}

Step-by-step explanation:

\star Let the width of a rectangular garden be 'w'

\star Length of a rectangular garden = 6 + w

\star Perimeter of a rectangular garden = 32 feet

<u>\longrightarrow</u><u> </u><u>Finding </u><u>the</u><u> </u><u>width</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangular</u><u> </u><u>garden</u> :

\boxed{ \sf{ \: Perimeter \: of \: a \: rectangle \:  =  \: 2(length + width}}

\mapsto{ \text{32 = 2(6 + w + w)}}

\text{Step \: 1 \:  : Collect \: like \: terms}

Like terms are those which have the same base

\mapsto{ \text{32 = 2(6 + 2w) }}

\text{Step \: 2 \:  : Distribute \: 2 \: through \: the \: parentheses}

\mapsto{ \text{32 = 12 + 4w}}

\text{Step \: 3 \:  : Swap \: the \: sides \: of \: the \: equation}

\mapsto{ \text{4w + 12 = 32}}

\text{Step \: 4 \:  : Move \: 12 \: to \: right \: hand \: side \: and \: change \: its \: sign}

\mapsto{ \text{4w = 32 - 12}}

\text{Step \: 5 \:  : Subtract \: 12 \: from \: 32}

\mapsto{ \sf{4w = 20}}

\text{Step \: 6 \:  : Divide \: both \: sides \: by \: 4}

\mapsto{ \sf{ \frac{4w}{4}  =  \frac{20}{4}}}

\text{Step \: 7 \:  : Calculate}

\mapsto{ \text{w = 5}}

Width of a rectangular garden = 5 feet

\longrightarrow <u>Substituting </u><u>/</u><u> </u><u>Replacing </u><u>the </u><u>value </u><u>of </u><u>w </u><u>in </u><u>6</u><u> </u><u>+</u><u> </u><u>w </u><u>in </u><u>order </u><u>to </u><u>find</u><u> </u><u>the</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangular </u><u>garden</u>

\mapsto{ \sf{Length = 6 + w = 6 + 5 =  \bold{11 \: feet} }}

\longrightarrow <u>Finding</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>rectangular</u><u> </u><u>garden</u><u> </u><u>having</u> <u>length of 11 feet and</u><u> </u><u>width</u><u> </u><u>of</u><u> </u><u>5</u><u> </u><u>feet</u> :

\boxed{ \sf{Area \: of \: a \: rectangle = length \:∗ \:  width}}

\mapsto{ \text{Area \:  =  \: 11 \: ∗ \:   5}}

\mapsto{ \sf{Area \:  = 55 \:  {ft}^{2} }}

Area of a rectangular garden = 55 ft²

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

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Anna35 [415]
Answer: A
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zysi [14]
X=-9

There is no slope in either equation. Therefore, x=-11 and x=-9 will be two vertical and parallel lines at (-11,0) and (-9,0)
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3 years ago
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nordsb [41]

Step-by-step explanation:

x= -2,

(x - 5) / 2= -6

( (-2) - 5) / 2= -6

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(2 - 5) / 2= -6

(-3) / 2= -6

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right hand side not equal to left hand side of this equation.so,x= 2 cannot exist for this equation.

x= -17

(x - 5) / 2= -6

( ( -17) - 5) / 2= -6

(- 22) / 2= -6

-11 = -6

right hand side not equal to left hand side of this equation.so,x= -17 cannot exist for this equation.

x= -7,

(x - 5) / 2= -6

( ( -7) -5) / 2= -6

(-12) / 2= -6

-6= -6

right hand side equal to left hand side of this equation.so,x= -7 exist for this equation.

8 0
3 years ago
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