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Serggg [28]
3 years ago
6

Two similar polygons have areas of 16 square inches and 64 square inches. The ratio of a pair of corresponding sides is 1/4.

Mathematics
1 answer:
Luden [163]3 years ago
5 0
Your answer is A. True.

Hope this helps.


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Need answer fast as possible
k0ka [10]

Answer:15

Step-by-step explanation:

The box on the left is square in a square all side are equal

7 0
3 years ago
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What is the area of the trapezoid to the nearest tenth? <br> PICTURE BELOW ⬇⬇⬇⬇
Nastasia [14]

Answer:

  62.4 ft²

Step-by-step explanation:

The unmarked horizontal dimension at the bottom of the triangle is ...

  (8 ft)sin(30°) = 4 ft

The unmarked vertical dimension of the triangle (the height of the trapezoid) is ...

  (8 ft)cos(30°) ≈ 6.93 ft

Then the area of the trapezoid is given by the formula ...

  A = (1/2)(b1 +b2)h

  A = (1/2)((4 ft+7 ft) +(7 ft))(6.93 ft) ≈ 62.4 ft²

_____

The mnemonic SOH CAH TOA can remind you of the relationships between right triangle dimensions and angles.

  Sin = Opposite/Hypotenuse   ⇒   Hypotenuse×Sin = Opposite

  Cos = Adjacent/Hypotenuse   ⇒   Hypotenuse×Cos = Adjacent

6 0
3 years ago
JIMMMMMM
Vitek1552 [10]
>= means greater than or equal to
<= means less than or equal to

---------------------------------------------

Part A

The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.

The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)

--------------

The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.

The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)

--------------

If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions. 

---------------------------------------------

Part B

Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not

For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.

Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3

We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas. 

7 0
3 years ago
Read 2 more answers
Solve:m+ 2/3=1/2 <br><br> A. 1 1/6<br><br> B. 1/6<br><br> C.-1<br><br> D. - 1/6
natita [175]

Answer:

D. - 1/6

Step-by-step explanation:

M + 2/3 = 1/2

M = 1/2 - 2/3

\frac{1}{2} - \frac{2}{3} = \frac{3}{6} -  \frac{4}{6} = -\frac{1}{6}

6 0
3 years ago
Read 2 more answers
The equation of the graph shown plotted in the coordinate plane is y = x + 4. What statement is not true about this graph? A. Ea
frez [133]

Definately option C

  • As it's mentioned y=x+4 it must be function
  • So in function every domain has its unique range.
  • Hence here the slope is not constant

As

The graph shown is a parabola which is not y=x+4

So option C is wrong

7 0
2 years ago
Read 2 more answers
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