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2 years ago

Como calcula-se a àrea de um cubo

Mathematics

1 answer:

Reil [10]2 years ago

3 0

Answer:

Spanish:

En palabras, el área de la superficie de un cubo es el área de los seis cuadrados que lo cubren. El área de uno de ellos es un * a, o un 2. Como todos son iguales, puedes multiplicar uno de ellos por seis, por lo que el área de la superficie de un cubo es 6 veces uno de los lados al cuadrado.

Además, un cubo tiene 6 caras. El área de cada cara se encuentra elevando al cuadrado la longitud del lado. Multiplica el área de una cara por el número de caras para obtener la superficie total del cubo.

Step-by-step explanation:

English:

In words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

Also, A cube has 6 faces. The area of each face is found by squaring the length of the side. Multiply the area of one face by the number of faces to get the total surface area of the cube.

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3 years ago

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A tree's height increase by 15%. If a tree is 23 feet high, how tall will it be in one year ?

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Increased height:-

$\\ \sf\longmapsto 23\times 15\%$

$\\ \sf\longmapsto 23\times \dfrac{15}{100}$

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2 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$