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2 years ago

At the playground, the new sandbox was 10 meters wide and had an area of 60 square meters. How long is the sandbox?

Mathematics

1 answer:

Gnoma [55]2 years ago

7 0

Answer:

6 meters

Step-by-step explanation:

Area is length times width. If the width is 10 and the area is 60, then you would find 10 times ??? equals 60. Or 60 divided by 10. which is 6

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En una tienda de ropa vas a comprarte una pantalones que están de 150 a 90 euros y una camiseta de 40 a 24 ¿ tienen el mismo por

uranmaximum [27]

Respuesta:

Sí, tienen el mismo descuento del 40%

Explicación paso a paso:

Dado :

PANTALÓN :

Precio inicial = 150

Precio con descuento = 90

Porcentaje de descuento en el precio:

(150 - 90) / 150 * 100%

60/150 * 100%

0,4 * 100

= 40%

CAMISA:

Precio inicial = 40

Precio con descuento = 24

Porcentaje de descuento en el precio:

(40-24) / 150 * 100%

16/40 * 100%

0,4 * 100

= 40%

Sí, tienen el mismo descuento del 40%

8 0

3 years ago

Write down a multiple of 8 that is between 20 and 50.

serg [7]

Answer:

24, 32, 40, or 48

Step-by-step explanation:

Count by 8.

8 0

3 years ago

Read 2 more answers

Lita jogged for 7*3/10 kilometers on Friday. She jogged 1*3/4 kilometres more on Saturday. How many kilometres did she jog on bo

Elza [17]

7*3/10 = 2.1

1*3/4 = 0.75

2.1 + 0.75 = 2.85

hope this helps :)

8 0

3 years ago

Area of square tile, math question

Inessa05 [86]

Answer:

Answer: √49

___________

√49 = 7

Step-by-step explanation:

The area of a square is equal to the square of its side or the product of its side and its side.

A = s², because the sides of a square are congruent.

5 0

3 years ago

For what value of k will the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent?

konstantin123 [22]

Answer:

After solve the equations we get value of k=3

Step-by-step explanation:

We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.

If the lines are concurrent, they pass through same point.

Let:

$x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)$

First solving equation 1 and 2 to find values of x and y

From eq(1) we find value of x and put it in eq(2)

$From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1$

After solving we get value of y=-1

Now putting in eq(1) to get value of x

$x+2y=0\\x+2(-1)=0\\x-2=0\\x=2$

So, Value of x= 2

Now put value of x=2 and y=-1 into eq(3) to find value of k

$5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3$

So, After solve the equations we get value of k=3

4 0

3 years ago