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jasenka [17]
3 years ago
13

Leo le debe 15.00 pesos. Si su madre le da 60.00 pesos de paga y se gasta 20.00 con sus amigos

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
7 0

Answer:

Le quedan 25 pesos

No tendrá pendiente ninguna deuda

Step-by-step explanation:

60-15-20=25

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Podcza

7 0
3 years ago
James joins Club One which charges a monthly membershi[ of $ 19.99. How much will James spend in all, if he continues his member
maria [59]

Answer:

James will spend $119.94

Step-by-step explanation:

Multiply $19.99 by 6.

<em>$19.99 x 6 = $119.94</em>

8 0
3 years ago
A farmer Has 900 yards of fencing to enclose a rectangular pasture for her goats. Since one side of the pasture borders a river,
horsena [70]

Answer:

Length, l=450 yards

Width, b=225 yards

Step-by-step explanation:

Let the long side be x and since the length is twice the width, then the width is half of that length hence 0.5x. Since we only have three parts to fence as one long side doesn't require fencing, then the perimeter will be given by x+0.5x+0.5x=2x=900 yards

X=900/2=450 yards for long side.

The short side is 0.5x hence 0.5*450=225 yards

In conclusion, the long side is 450 yards while short side is 225 yards.

7 0
3 years ago
What is the value of y in the solution to the system of equations?
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Step-by-step explanation:

8 0
3 years ago
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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
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