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Nastasia [14]
3 years ago
9

PLZ HELP PPLLLZZZZ ASAP PLZ HELP PPPPLLLLZZZZZZ PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPLLLLLLLLLLLLLLLLLLLLLLLLLLL

LLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

56 m^2

Step-by-step explanation:

a= b x h

= 7 x 8

= 56m^2

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What is the value of 12 (8 + 9)?<br> 105<br> 204<br> 357<br> 864
USPshnik [31]

Answer:

204

Step-by-step explanation:

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3 0
3 years ago
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Solve for x in this figure​
Alecsey [184]

The sum of the angles in a pentagon is 540°

540 - (93 + 90 + 109 + 132)

540 - (424) = 116

x would be 116°

8 0
3 years ago
A cube has a volume of 1/8 cubic meter. Complete the explanation for finding the length of each side of the cube. Express your a
musickatia [10]

Answer:

Volume of a cube= 1/8  cubic meter

We know that volume of a cube= side x side x side

(side)^{3}= \frac{1}{8}

From here, we get side of cube= \frac{1}{2} meter

Hence, the side of a cube =     \frac{1}{2} meter

We have to find a fraction, when it is multiplied by itself

\frac{1}{2} x \frac{1}{2}  = \frac{1}{4}

Here we get our answer  \frac{1}{4}

8 0
3 years ago
The proof, with a missing reason, proves that the measure of angle ECB is 54°.
aniked [119]

Answer:

Midsegment of a Triangle Theorem

Step-by-step explanation:

In\: \triangle ADE\\m\angle A + m\angle D + m\angle AED = 180°\\\therefore 90°+36°+m\angle AED= 180°\\\therefore 126° +m\angle AED = 180°\\\therefore m\angle AED = 180°-126°\\\therefore m\angle AED = 54°\\

In\: \triangle ADE, D and E are mid points of sides AB and AC respectively.

Therefore, DE || BC

Hence, by Midsegment of a Triangle Theorem

m\angle ACB = m\angle AED =54°(corresponding\:\angle 's) \\\therefore m\angle ECB = m\angle AED =54°(\because A - E - C)

3 0
3 years ago
quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2). Which statement is true about the dilation?
Setler [38]

Answer:

The vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

Step-by-step explanation:

<em>Note: You missed to mention the vertices of a quadrilateral ABCD. So, I am assuming the following vertices. It would anyways clear your concept.</em>

<em />

Let us suppose

  • A(-5, 1)
  • B(-4, 3)
  • C(1, 2)
  • D(-3, 0)

We know that the rule of the dilation with the center of dilation at (2,2) by a scale factor of 2 is:

  • (x, y) → (2x-2, 2y-2)

Therefore, the vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

6 0
3 years ago
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