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2 years ago

14

Cual es el resultado de 12/6?

2 answers:

Sati [7]2 years ago

6 0

Answer:

2

Step-by-step explanation:

oksano4ka [1.4K]2 years ago

3 0

Answer:

Resultado es dos (2)

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Tanya has a garden with a trench around it. The garden is a rectangle with length 2.5m and width 2m. The trench and the garden t

Drupady [299]

Answer:

5.5m^2

the area of just the trench is 5.5 m^2

Step-by-step explanation:

The Area of trench only is the total area(trench and garden) minus the Area of garden

Area of trench only = total area(trench and garden) - Area of garden

Area = length × breadth

Substituting the given dimensions;

Total area = 3.5 × 3 = 10.5 m^2

Area of garden = 2.5×2 = 5 m^2

Area of trench only = 10.5 m^2 - 5 m^2

Area of trench only = 5.5m^2

the area of just the trench is 5.5 m^2

8 0

3 years ago

Read 2 more answers

15pts please help asap<br> 2.5 as an improper fraction?

RideAnS [48]

2.5 = 2 5/10 as a Mixed fraction

To get a improper fraction, multiply the whole number with the denominator, and then add the numerator to the answer, giving the numerator (keep the denominator number)

2 x 10 = 20
20 + 5 = 25

25/10 is your answer

hope this helps

6 0

3 years ago

What is the answer to (-7)(2)(-2).

IgorC [24]

Answer:

28

Step-by-step explanation:

-2 × -7 = 14

14 × 2 = 28

thats the correct answer

8 0

3 years ago

Catherine saved some money and plans to add the same amount each week to her savings account. The table represents the number of

yan [13]

Answer: y=6x+50 B

Step-by-step explanation:

5 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers