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pantera1 [17]
3 years ago
5

 which side lengths form a right triangle

Mathematics
2 answers:
katrin [286]3 years ago
4 0

Answer:

I think b

Step-by-step explanation:

m_a_m_a [10]3 years ago
3 0

Answer:

B

Step-by-step explanation:

You might be interested in
What is the perimeter of a square which has the same area as a circle with circumference of 4π?
enot [183]

Answer:

perimeter = 8\sqrt{\pi }

Step-by-step explanation:

We require to calculate the area (A) of the circle

A = πr² ← r is the radius

circumference = 4π = 2πr, hence

2πr = 4π ( divide both sides by 2π )

r = 2, hence

A = π × 4 = 4π

The area of the square is therefore 4π and area = s² ← s is the side length

s² = 4π ( take the square root of both sides )

s = \sqrt{4\pi } = 2\sqrt{\pi }, hence

perimeter = 4 × 2\sqrt{\pi } = 8\sqrt{\pi }



8 0
3 years ago
In a school presidential race, Jeb received 100 votes, Janine 60, and Arnie 40 votes. Make a circle graph to display the data as
Anarel [89]

Answer:

Jeb: 50%, Janine: 30%, Arnie: 20%

Step-by-step explanation:

100+60+40=

200 votes in total

1% of 200 = 2

100 divided by 2 = 50

60 divided by 2 = 30

40 divided by 2 = 20

Hope this helps!

7 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!
amid [387]

Answer:

Lauren is 19. Lauren's grandmother is 61.

Step-by-step explanation:

19 * 3 = 57

57 + 4 =61

19 + 61 = 80

7 0
3 years ago
Read 2 more answers
Are the graphs of y= 1/3x-4 and y-2=-3(x-4) parallel, perpendicular, or neither
melamori03 [73]

Answer:

ghtrh

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1.Encuentra la ecuación de la circunferencia que tiene como ecuación ordinaria (x-4)² + (y+5)²= 25 2- encuentra la ecuación gene
rodikova [14]

Responder:

1) x² + y²-8x + 10y + 16 = 0

2) x² + y² + x + 18y + 33 = 0

Explicación paso a paso:

A partir de la pregunta, debemos expresar la ecuación en la forma estándar de un círculo expresado como;

x² + y² + 2gx + 2fy + c = 0 donde;

(-g, -f) es el centro.

1) Para la ecuación (x-4) ² + (y + 5) ² = 25

Expande el paréntesis;

(x-4) ² + (y + 5) ² = 25

x²-8x + 16 + y² + 10x + 25 = 25

x²-8x + 16 + y² + 10y + 25-25 = 0

Recopile términos similares;

x² + y²-8x + 10y + 16 + 0 = 0

<em>x² + y²-8x + 10y + 16 = 0 </em>

<em>Por lo tanto, la ecuación requerida es x² + y²-8x + 10y + 16 = 0 </em>

<em> </em>

2) Para la expresión (x + 1) ² + (y + 9) ² = 49

x² + x + 1 + y² + 18y + 81 = 49

x² + x + 1 + y² + 18y + 81-49 = 0

Recopile términos similares;

x² + y² + x + 18y + 82-49 = 0

<em>x² + y² + x + 18y + 33 = 0 </em>

<em>Por tanto, la ecuación requerida es x² + y² + x + 18y + 33 = 0</em>

4 0
2 years ago
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