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3 years ago

What does a supplementary angle look like

Mathematics

1 answer:

Keith_Richards [23]3 years ago

8 0

Answer:

It has no special appearance.

Step-by-step explanation:

Any angle of measure 180° or less is supplementary to some angle. A supplementary angle is one that is the difference between 180° and the angle you have. That is, two supplementary angles total 180°.

Supplementary angles are readily identifiable in a number of geometries. Adjacent angles of a parallelogram are supplementary; linear angles are supplementary. Same-side interior angles where a transversal crosses parallel lines are supplementary.

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In the diagram below, m_MON = 1249, m _ MOP = 2x + 1, and m2 NOP = 2x

miv72 [106K]

Answer:

The answer is below

Step-by-step explanation:

The Angle Addition Postulate states that the measure of an angle formed by two or more angles which are placed side by side is the sum of the measures of the two angles.

Therefore:

∠MON = ∠MOP + ∠NOP (angle addition postulate)

Substituting values gives:

124 = (2x + 1) + (2x + 1)

124 = 2x + 2x + 1 + 1

124 = 4x + 2

subtracting 2 from both sides of the equation:

124 - 2 = 4x + 2 - 2

4x = 122

Dividing through by 4:

4x / 4 = 122 / 4

x = 30.5

Therefore ∠MOP = 2x + 1 = 2(30.5) + 1 = 62°, ∠NOP = 2x + 1 = 2(30.5) + 1 = 62°

∠MOP = 62°, ∠NOP = 62°

5 0

3 years ago

What's 950 dividend by 48

stepladder [879]

19.79 should be the correct answer?

4 0

3 years ago

Read 2 more answers

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

What type of polynomial is this

Crank

Answer:

trinomial..............

5 0

3 years ago

Whats greater -4 or -10

anygoal [31]

-4 because it is closer to the 0

5 0

3 years ago

Read 2 more answers