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madreJ [45]
3 years ago
5

I NEED HELP ASAP LIKE NOWWW

Mathematics
1 answer:
jeka943 years ago
5 0

Answer:

C / (2pi) = r;  r= 13.9 cm

2A/b = h   ;  h= 5

Step-by-step explanation:

The formula for the circumference of a circle is  C = 2* pi * r.

Solving for r, we need to divide by 2 * pi on each side

C/(2* pi) = 2*pi*r/ (2* pi)

C / (2pi) = r

If the circumference is 87, we can find r by substituting into the equation.

87 / (2 * 3.14) = r

87/ 6.28 = r

13.85350318=r

Rounding to the nearest tenth

r= 13.9 cm


The formula for the area of a triangle is  A = bℎ /2  

To solve the formula for h, we need to multiply each side by 2

2*A = bh

Then divide each side by b

2A/b = bh/b

2A/b = h


If the  area  of a triangle is 25  and the base of 10 , we can substitute in to find the height.

2* 25 /10 = h

50/10 = h

5 =h

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7 0
3 years ago
The vertices of Quadrilateral ABCD are located at (1, 4), (5, 0), (2, –3), and (–2, –2).
forsale [732]

Answer:

A - Rectangle B - Square

C - Parallelogram D - Rhombus

Explanation:

We are given

A

(

1

,

2

)

,

B

(

2

,

−

2

)

and hence

A

B

=

√

(

2

−

1

)

2

+

(

−

2

−

2

)

2

=

√

17

. Further slope of

A

B

is

−

2

−

2

2

−

1

=

−

4

1

=

−

4

.

Case A -

C

(

−

6

,

−

4

)

,

D

(

−

7

,

0

)

As

C

D

=

√

(

−

7

−

(

−

6

)

)

2

+

(

0

−

(

−

4

)

)

2

=

√

17

and slope of

C

D

is

0

−

(

−

4

)

−

7

−

(

−

6

)

=

4

−

1

=

−

4

As

A

B

=

C

D

and

A

B

||

C

D

slopes being equal, ABCD is a parallelogram.

graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+6)^2+(y+4)^2-0.08)((x+7)^2+y^2-0.08)=0 [-10, 10, -5, 5]}

Case B -

C

(

6

,

−

1

)

,

D

(

5

,

3

)

As

C

D

=

√

(

5

−

6

)

2

+

(

3

−

(

−

1

)

)

2

=

√

17

and slope of

C

D

is

0

−

(

−

4

)

−

7

−

(

−

6

)

=

4

−

1

=

−

4

Further,

B

C

=

√

(

6

−

2

)

2

+

(

−

1

−

(

−

2

)

)

2

=

√

17

and slope of

B

C

is

−

1

−

(

−

2

)

6

−

2

=

1

4

As

B

C

=

A

B

and they are perpendicular (as product of slopes is

−

1

), ABCD is a square.

graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-6)^2+(y+1)^2-0.08)((x-5)^2+(y-3)^2-0.08)=0 [-10, 10, -5, 5]}

Case C -

C

(

−

1

,

−

4

)

,

D

(

−

2

,

0

)

As mid point of

A

C

is

(

1

−

1

2

,

2

−

4

2

)

i.e.

(

0

,

−

1

)

and midpoint of

B

D

is

(

2

−

2

2

,

−

2

+

0

2

i.e.

(

0

,

−

1

)

i.e. midpoints of

A

C

and

B

D

are same,

but,

B

C

=

√

(

2

−

(

−

1

)

)

2

+

(

−

2

−

(

−

4

)

)

2

=

√

13

i.e.

A

B

≠

B

C

and hence ABCD is a parallelogram.

graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x+2)^2+y^2-0.08)=0 [-10, 10, -5, 5]}

Case D -

C

(

1

,

−

6

)

,

D

(

0

,

−

2

)

As mid point of

A

C

is

(

1

+

1

2

,

2

−

6

2

)

i.e.

(

1

,

−

2

)

and midpoint of

B

D

is

(

2

+

0

2

,

−

2

+

(

−

2

)

2

i.e.

(

1

,

−

2

)

i.e. midpoints of

A

C

and

B

D

are same,

and,

B

C

=

√

(

2

−

1

)

2

+

(

−

2

−

(

−

6

)

)

2

=

√

17

i.e.

A

B

=

B

C

and hence ABCD is a rhombus.

graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-1)^2+(y+6)^2-0.08)(x^2+(y+2)^2-0.08)=0 [-14, 14, -7, 7]}

3 0
3 years ago
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