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worty [1.4K]
3 years ago
7

*** PLS HELP ASAP (14 points)

Mathematics
2 answers:
serious [3.7K]3 years ago
8 0

Answer:

Option 3. 360°

Step-by-step explanation:

we know that

The sum of the measures of the interior angles of triangle is equal to 180 degrees

The figure has two triangles

therefore

2(180\°)=360\°

<em>Alternative Method</em>

The formula to calculate the sum of the measures of the interior angles of a polygon is equal to

S=180(n-2)

where

n is the number of sides of the polygon

In this problem we have a a quadrilateral

so

n=4

substitute

S=180(4-2)

S=180(2)

S=360\°

mr Goodwill [35]3 years ago
8 0

Answer:

C: 360 degrees

Step-by-step explanation:

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A similar scale model of a building is 12 in tall and 9 in wide. If the actual building is 100 ft tall, how wide is the building
Dmitry [639]
You just put the scale of one building on one side and the scale of the similar building on other side, with width and length written in same place (denominator or numerator). With x for the part that you dont know

Length of one building / width of one building = length of other building / width of other building

12 / 9 = 100 / x

Solve for x, by cross-multiplying

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4 0
2 years ago
Solve the system for x and y -3x+8y=103 -5x-8y=-255
Romashka-Z-Leto [24]

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Answer:

  (x, y) = (19, 20)

Step-by-step explanation:

The coefficients of y are opposites, so we can add the two equations to eliminate y.

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  -8x = -152

  x = 19 . . . . . . . divide by -8

Substituting into the first equation, we have ...

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The solution is (x, y) = (19, 20).

3 0
3 years ago
Does 4(2x-5) equal 8x-5 or 8x-20? explain
Scorpion4ik [409]
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8 0
3 years ago
Read 2 more answers
Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
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3 years ago
Plzzz guys, help me with this​
babunello [35]

Answer:

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3 years ago
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