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2 years ago

Calculate cos 0 to two decimal places

Mathematics

2 answers:

pashok25 [27]2 years ago

7 0

For this case we have by definition of the Cosines Law that:

$7 ^ 2 = 10 ^ 2 + 8 ^ 2-2 (10) (8)$ * cosΘ

$49 = 100 + 64-160$ cosΘ

$49 = 164-160$ cosΘ

$49-164 = -160$ cosΘ

$-115 = -160$ cosΘ

$115 = 160$ cosΘ

cosΘ= $\frac {115} {160} = 0.71875$

Rounding off we have, 0.72

ANswer:

Option C

Sergio [31]2 years ago

3 0

We can use the law of cosines as follows:

$7^2 = 8^2+10^2-2\cdot 8 \cdot 10 \cdot \cos(\theta)$

We can rewrite this equation as

$49 = 164-160 \cdot \cos(\theta) \iff 160 \cdot \cos(\theta) = 115 \iff \cos(\theta)=\dfrac{115}{160}\approx 0.72$

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An ant begins at the top of the pictured octahedron. If the ant takes two "steps", what is the probability it ends up at the bot

Aleksandr-060686 [28]

Answer:

$P_{bottom}=\frac{1}{4}=0.25$

Step-by-step explanation:

Starting from the top, the ant can only take four different directions, all of them going down, every direction has a probability of 1/4. For the second step, regardless of what direction the ant walked, it has 4 directions: going back (or up), to the sides (left or right) and down. If the probability of the first step is 1/4 for each direction and once the ant has moved one step, there are 4 directions with the same probability (1/4 again), the probability of taking a specific path is the multiplication of the probability of these two steps:

$P_{2steps}=\frac{1}{4}*\frac{1}{4}=\frac{1}{16}$

There are only 4 roads that can take the ant to the bottom in 2 steps, each road with a probability of 1/16, adding the probability of these 4 roads:

$P_{bottom}=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

The probability of the ant ending up at the bottom is $\frac{1}{4}$ or 0.25.

6 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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2 years ago

Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 an

Yuri [45]

Answer:

For the given explanation we see that two life times are not independent

Step-by-step explanation:

probability for X (for x≥ 0)

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dy$

$-e^-^x\int\limits^\infty_0 {de^-^x^y} \,$

e⁻ˣ

Probability for X exceed 3

= $\int\limits^\infty_3 {f(x)dx$

= $\int\limits^\infty_3 {e^-^3 dx$

= $e^-^3$

probabilty for y≥ 0 is

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dx \\$

$\int\limits^\infty_0{-x/1+yd(e^-^x^(^1^+^y^))} \, =1/(1+y)² \\$

3 0

2 years ago

The two solids are similar and the ratio between the lengths of their edges is 2:9. What is the ratio of their surface areas? A)

masya89 [10]

When tow solids of similar shape that is assuming to be a box in this case, having a length ratio of 2:9, we get the ratio of the surface areas equal to thesquare of the ratio of their edges. This is becaue the surface area is equal to the area of the base (square) which is the square of the side. Answer is B.

7 0

2 years ago

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Solve the system using elimanation: 5x + 4y = 12. 3x -3y = 18

Scorpion4ik [409]

If you would like to solve the system of equations 5x + 4y = 12 and 3x - 3y = 18 using elimination, you can do this using the following steps:

5x + 4y = 12 /*3
3x - 3y = 18 /*4
_____________
15x + 12y = 36
12x - 12y = 72
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15x + 12y + 12x - 12y = 36 + 72
15x + 12x = 36 + 72
27x = 108
x = 108/27
x = 4

3x - 3y = 18
3 * 4 - 3y = 18
12 - 3y = 18
12 - 18 = 3y
3y = -6
y = -6/3
y = -2

(x, y) = (4, -2)

The correct result would be: x = 4 and y = -2.

4 0

2 years ago

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