Cuánto es la mitad de cinco octavos

1 answer:

4 0

Mitad de $\frac{5}{8}'s$
Multiplicar $\frac{5}{8}$ por $\frac{1}{2}$
$(\frac{5}{8}) (\frac{1}{2}) = \frac{(5)(1)}{(8)(2)} = \frac{5}{16}$

You might be interested in

I have a question for someone out there I'm 10 and need some help what is 1/10+10/100

Ratling [72]

Answer:

0.2

Step-by-step explanation:

1/10 = 0.1

10/100 = 0.1

0.1+0.1 = 0.2

7 0

3 years ago

What's the estimate for 2-9

andrew11 [14]

The estimate between 2 and 9 is 7

5 0

3 years ago

A farmer has a field in the shape of a triangle. The farmer has asked the manufacturing class at your school to build a metal fe

Finger [1]

Answer:

Fencing required = 1586 m

Step-by-step explanation:

The given statements can be thought of a triangle $\triangle ABC$ as shown in the diagram attached.

A be the 1st vertex, B be the 2nd vertex and C be the 3rd vertex.

Distance between 1st and 2nd vertex, AB = 435 m

Distance between 2nd and 3rd vertex, AC = 656 m

$\angle A =49^\circ$

To find:

Fencing required for the triangular field.

Solution:

Here, we know two sides of a triangle and the angle between them.

To find the fencing or perimeter of the triangle, we need the third side.

Let us use Cosine Rule to find the third side.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\Rightarrow cos 49^\circ = \dfrac{656^{2}+435^{2}-BC^{2}}{2\times 435\times 656}\\\Rightarrow BC^2 = 430336+189225-2 (435)(656)cos49^\circ\\\Rightarrow BC^2 = 430336+189225-570720\times cos49^\circ\\\Rightarrow BC^2 =619561-570720\times cos49^\circ\\\Rightarrow BC \approx 495\ m$