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raketka [301]
3 years ago
11

According to the distributive property, 6 (5 + 3) =

Mathematics
2 answers:
notsponge [240]3 years ago
8 0
D.
6 (5+3) means six times five plus six times three
miv72 [106K]3 years ago
6 0
A, but im not 100% it could be B

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All 3 questions 3 pictures for find the missing angles, please with reasoning why? 100 points
Katyanochek1 [597]

Answer:

Problem 1)

m\angle DEG = 38^\circ

Problem 2)

m\angle x =140^\circ \text{ and } m\angle y = 49^\circ

Problem 3)

m\angle EDG = 65^\circ

Step-by-step explanation:

Problem 1: ∠DEG

From the diagram, we know that ∠DFG intercepts Arc DG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=m\angle DFG

We know that m∠DFG = 38°. So:

\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ

∠DEG also intercepts Arc DG. Hence:

\displaystyle m\angle DEG=\frac{1}{2}m\stackrel{\frown}{DG}

We know that Arc DG measures 76°. Hence:

\displaystyle m\angle DEG =\frac{1}{2}\left(76\right)=38^\circ

Alternate Explanation:

Since ∠DEG and ∠DFG intercept the same arc, ∠DEG ≅ DFG. So, m∠DEG = m∠DFG = 38°.

Problem 2: Circle with Centre O

(Let the bottom left corner be A, upper be B, and right be C.)

∠ACB intercepts Arc AC.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=m \angle ACB

Since m∠ACB = 70°:

\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ

∠<em>x</em> is a central angle and also intercepts Arc AC.

The measure of a central angle is equal to its intercepted arc. Thus:

\displaystyle m\angle x =m\stackrel{\frown}{AC}=140^\circ

The sum of the interior angles of a polygon is given by the formula:

(n-2)\cdot 180^\circ

Where <em>n</em> is the number of sides.

Since the inscribed figure is a four-sided polygon, its interior angles must total:

(4-2)\cdot 180^\circ =360^\circ

Therefore:

21+70+y+(360-x)=360

Substitute and solve for <em>y:</em>

91+y+(360-140)=360\Rightarrow y +311=360\Rightarrow m\angle y = 49^\circ

Problem 3: ∠EDG

∠EFG intercepts Arc EDG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG}=m\angle EFG

Since m∠EFG = 115°:

\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG} = 115\Rightarrow m\stackrel{\frown}{EDG} = 230^\circ

A full circle measures 360°. Hence:

m\stackrel{\frown}{EDG}+m\stackrel{\frown}{EFG}=360^\circ

Since we know that Arc EDG measures 230°:

230^\circ + m\stackrel{\frown}{EFG}=360

Solve for Arc EFG:

m\stackrel{\frown}{EFG}=130^\circ

∠EDG intercepts Arc EFG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

\displaystyle m\angle EDG = \frac{1}{2}m\stackrel{\frown}{EFG}

Since we know that Arc EFG measures 130°:

\displaystyle m\angle EDG = \frac{1}{2}(130)=65^\circ

7 0
3 years ago
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