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3 years ago

11

What number is 12% of 90

2 answers:

yulyashka [42]3 years ago

7 0

10.8

12% = 10% + 1% + 1%

10 percent of a number is that number but with the decimal point 1 to the left. 10% of 90.0 is 9.0

1% of a number is that number but with the decimal point two to the left. 1% of 90.0 is 0.9

9 + 0.9 + 0.9 = 10.8

10.8 is 12% of 90

Tasya [4]3 years ago

3 0

Answer:

10.8

Step-by-step explanation:

i used a calculator

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XW must be congruent to YZ, and WZ must be congruent to XY

In order for it to be a parallelogram, the opposite sides must be congruent.

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3 years ago

X + 6 >10 or 3x + 11 ≤ −7

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Answer:

x>4 or x≤-6

Step-by-step explanation:

x+6>10 or 3x+11≤-7

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3 years ago

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Masteriza [31]

Answer:

(2,-5) and (2,1).

Step-by-step explanation:

We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.

A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.

Standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center and r is the radius.

$(x-(-2))^2+(y-(-4))^2=(5)^2$

$(x+2)^2+(y+2)^2=25$

Now, put x=2.

$(2+2)^2+(y+2)^2=25$

$(4)^2+(y+2)^2=25$

$(y+2)^2=25-16$

$(y+2)^2=9$

Taking square root on both sides.

$y+2=\pm\sqrt{9}$

$y+2=\pm3$

$y+2=-3\text{ or }y+2=3$

$y=-5\text{ or }y=1$

So, the y-coordinates of the points are -5 and 1.

Therefore, the required points are (2,-5) and (2,1).

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3 years ago

En un aeropuerto dos aviones A1 y A1 se acercan para aterrizar, si la ecuación de la trayectoria del primero es -x+2y-100=0, mie

liraira [26]

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

-------------------------

La trayectoria del primero es:

$-x + 2y - 100 = 0$

$x_1 = 2y - 100$

Para el segundo, tiene-se que:

$-x + 200 + 2y = 0$

$x_2 = 2y + 200$

Igualando los valores de x:

$x_1 = x_2$

$2y - 100 = 2y + 200$

$0y = 300$

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

Un problema similar es dado en brainly.com/question/24653364

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schepotkina [342]

Answer:

47min.

Step-by-step explanation:

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3 years ago

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