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2 years ago

Fifteen times the square of a non zero number is equal to 15 times the number what is the number ?

1 answer:

3 0

1
because

1^2=1

1^2×15=15

1×15=15

So 1^2*15=1×15

Equation looks like this

15x^2=15x

subtract 15x

15x^2+15x=0

factor

15x(x-1)=0

solutions are 0 or 1 but seeing it said non zero number it must be 1

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2 years ago

A metal rod will be cut into pieces that are each 1/56 meters long. The rod is 7/8 meters long. How many pieces will be made fro

S_A_V [24]

The number of pieces made from rod is 49

<h3><u>Solution:</u></h3>

Given that A metal rod will be cut into pieces that are each 1/56 meters long

The rod is 7/8 meters long.

<em><u>To find: number of peices made from rod</u></em>

The number of peices made from rod can be calculated by dividing the total length of rod by length of each piece

$\text {number of pieces made }=\frac{\text {total length of rod }}{\text { length of each piece}}$

Here total length of rod = $\frac{7}{8}$ meters

length of each piece to be made = $\frac{1}{56}$ meters

Substituting the values in above formula, we get

$\text {number of pieces made }=\frac{\frac{7}{8}}{\frac{1}{56}}=\frac{7}{8} \times \frac{56}{1}$

$\text {number of pieces made }=7 \times 7=49$

So the number of pieces made from rod is 49

7 0

2 years ago

A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

Step-by-step explanation:

we know that

we can divide the regular hexagon into 6 identical equilateral triangles

see the attached figure to better understand the problem

The approximate area of the circle is approximately the area of the six equilateral triangles

Remember that

In an equilateral triangle the interior measurement of each angle is 60 degrees

We take one triangle OAB, with O as the centre of the hexagon or circle, and AB as one side of the regular hexagon

Let

M ----> the mid-point of AB

OM ----> the perpendicular bisector of AB

x ----> the measure of angle AOM

$m\angle AOM =30^o$

In the right triangle OAM

$tan(30^o)=\frac{(a/2)}{r}=\frac{a}{2r}\\\\tan(30^o)=\frac{\sqrt{3}}{3}$

$\frac{a}{2r}=\frac{\sqrt{3}}{3}$

we have

$r=3\ units$

substitute

$\frac{a}{2(3)}=\frac{\sqrt{3}}{3}\\\\a=2\sqrt{3}\ units$

Find the area of six equilateral triangles

$A=6[\frac{1}{2}(r)(a)]$

simplify

$A=3(r)(a)$

we have

$r=3\ units\\a=2\sqrt{3}\ units$

substitute

$A=3(3)(2\sqrt{3})\\A=18\sqrt{3}\ units^2$

Therefore

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$

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2 years ago