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Nitella [24]
3 years ago
8

A quadrilateral has no pairs of parallel sides.what two shapes could be it

Mathematics
2 answers:
Basile [38]3 years ago
8 0

Answer:

All trapezoids are quadrilaterals.

Step-by-step explanation:

Incorrect. Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.

saul85 [17]3 years ago
7 0

Answer:

i need help it is the wrong answer

Step-by-step explanation:

All trapezoids are quadrilaterals.

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9.0 - 0.178=<br>5 points<br>Your answer​
bagirrra123 [75]

Answer:

8.822. Hope this helps.

3 0
3 years ago
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Amir pits £3035 into a bank account. The account pays 4% compound interest each year. Work out how much money amir will have in
myrzilka [38]

Answer:

\£3840.2

Step-by-step explanation:

Let P denotes principal amount, T denotes time period and R denotes rate of interest.

Amount = P(1+\frac{R}{100})^T

Amir pits £3035 into a bank account. The account pays 4% compound interest each year.

Put P=\£3035,\,R=4\%,\,T=6

Amount = 3035(1+\frac{4}{100})^6=3035(\frac{104}{100})^6=\£3840.2

Therefore, Amir will have \£3840.2 in the account after 6 years.

7 0
3 years ago
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lisov135 [29]
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3 years ago
Work out the size of AED. Work out x
Monica [59]

Answer:

a). m∠AED = 70°

b). x = 10°

Step-by-step explanation:

a). Quadrilateral ABDE is a cyclic quadrilateral.

Therefore, by the theorem of cyclic quadrilateral,

Sum of either pair of opposite angle is 180°

m(∠AED) + m(∠ABD) = 180°

m(∠AED) = 180° - 110°

m(∠AED) = 70°

Since, ∠AED ≅ ∠EAD

Therefore, m∠AED = m∠EAD = 70°

b). By triangle sum theorem in ΔABD,

m∠ABD + m∠BDA + m∠DAB = 180°

110° + 40° + m∠DAB = 180°

m∠DAB = 180° - 150°

m∠DAB = 30°

m∠BAE = m∠EAD + m∠BAD

              = 70° + 30° = 100°

By angle sum theorem in ΔACE,

m∠EAC + m∠AEC + m∠ACE = 180°

100° + 70° + x° = 180°

x = 180° - 170°

x = 10°

4 0
3 years ago
HELPP!!what is the answer to this picture above^
nasty-shy [4]

Answer:

\fbox{ \fbox { \sf{Distance  =  \sf{15 \: units}}}}

{ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}

Step-by-step explanation:

\star{ \:  \sf { \: Let \: the \: points \: be \: A \: and \: B}}

\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}

\underline{ \underline{ \tt{Finding \: the \: distance}}}

\boxed{ \sf{Distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(18 - 6)}^{2}  +  {(15 - 6)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(12)}^{2}  +  {(9)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{144 + 81}}}

\mapsto{ \sf{Distance =  \sqrt{225} }}

\mapsto{ \sf{Distance =  \sqrt{ {15}^{2} } }}

\mapsto{  \boxed{\sf{Distance = 15 \: units}}}

\underline{ \underline {\tt{Finding \: the \: Midpoint}}}

\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2}  \: , \:  \frac{y1 + y2}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2}  \: , \:  \frac{6 + 15}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{24}{2}  \: , \:  \frac{21}{2} )}}

\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

3 0
3 years ago
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