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solong [7]
4 years ago
7

The weight of an object on the moon is 1/6 its weight on earth. write 1/6 as a decimal

Mathematics
2 answers:
denpristay [2]4 years ago
7 0
Point six repeatedly so 0.16666666666... so just put a bar over the six 
san4es73 [151]4 years ago
3 0

Answer:

Weight of the object on the moon

      =\frac{1}{6} \times \text{Weight of an object on the earth}

If in a fraction,that is rational number, denominator has term other than

                                   2^a , \text{or},5^b,\text{or} ,2^a \times 5^b}

then the decimal expansion will be non terminating repeating.

 \rightarrow\frac{1}{6}=0.1666.....\\\\=0.167\text{approx}

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The fraction models below represent two fractions of the same whole. How much of the Fraction 5 over 8 is in the Fraction 1 over
IrinaVladis [17]

To find how much of 5/8 is you would need to divide 1/2 by 5/8:

X = (1/2) / (5/8)

When dividing fractions, you reverse the second fraction and multiply:

X = (1/2) x (8/5)

X = 8/10

Simplify:

X = 4/5

5 0
3 years ago
Read 2 more answers
Hello can anyone help me plz its math
AlladinOne [14]
The answer would be 4/9 hope it helped(:
3 0
3 years ago
NEED ASAP!!!<br> 1. Find the value of x in the diagram below.<br> 120°<br> x
TiliK225 [7]

x and the 120 degrees are Complementary angles so they are the same. x=120˚

6 0
3 years ago
How do I solve -0.4r=16?
Leviafan [203]

Answer:

r = -40

Step-by-step explanation:

1. Move constants to the other side of the equation

r = 16/-.04

2. Simplify (by calculator)

You can use a calculator and plug in 16/-.04

r = -40

2. Simplify (by fraction)

You can turn -.4 into a fraction.

-.04 = - 4/10

r = 16/-(4/10)

r = 16*10/-4

r = -40

6 0
3 years ago
consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros
Bezzdna [24]

Answer:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

Step-by-step explanation:

The given quadratic form is of the form

q(x,y,z)=ax^2+by^2+dxy+exz+fyz.

Where a=11,b=-1,c=-4,d=-16,e=8,f=-4.Every quadratic form of this kind can be written as

q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)

Observe that A is a symmetric matrix. So A is orthogonally diagonalizable, that is to say,  D=Q^{T}AQ where Q is an orthogonal matrix and D is a diagonal matrix.

In our case we have:

A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)

The eigenvalues of A are \lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5.

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)

D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)

Now, we have to do the change of variables {\bf x}=Q{\bf y} to obtain

q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2

Which can be written as:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

4 0
4 years ago
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