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Setler [38]
3 years ago
12

D. 5

Mathematics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

skdjdnfjtnthhr potlas

rmfmrmfmfmrmrjrjfk fkdjtmgjggmjt ne jfjdckmcjfjfjfjgjgmgngjgjv. vb. hahaha

eorirififjtjrjrjrjejfjrjfjrjfjfjfnfmc h için mvnvnffnfnnfdmm

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Five students are competing in a race. In how many different
Dimas [21]

Answer:

120

Step-by-step explanation:

• we need to find the number of the possible orderings

of this five students.

•• which we can simulate to a permutation

of 5 objects.

••• and we know that The number of permutations

of 5 distinct objects is 5!.

5! = 5×4×3×2×1 = 120

there are six permutations of the set {1, 2, 3}, namely (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). These are all the possible orderings of this three-element set.

The number of permutations of n distinct objects is n factorial, usually written as n!, which means the product of all positive integers less than or equal to n.

6 0
2 years ago
<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%20%3D%200" id="TexFormula1" title=" {x}^{2} + 2x = 0" alt=
Alex

Answer:

\textbf{Hello!}

\Longrightarrow2^2+2z=0

\Longrightarrow x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \:0}}{2\cdot \:1}

\Longrightarrow \sqrt{2^2-4\cdot \:1\cdot \:0}

\Longrightarrow =\sqrt{2^2-0}

\Longrightarrow =\sqrt{2^2}

\Longrightarrow=2

\Longrightarrow x_{1,\:2}=\frac{-2\pm \:2}{2\cdot \:1}

\Longrightarrow x_1=\frac{-2+2}{2\cdot \:1},\:x_2=\frac{-2-2}{2\cdot \:1}

\Longrightarrow\frac{-2+2}{2\cdot \:1}

\Longrightarrow =\frac{0}{2\cdot \:1}

\Longrightarrow =\frac{0}{2}

=0

\Longrightarrow\frac{-2-2}{2\cdot \:1}

\Longrightarrow =\frac{-4}{2\cdot \:1}

\Longrightarrow =\frac{-4}{2}

\Longrightarrow =-\frac{4}{2}

=-2

x=0,\:x=-2\Longleftarrow

\underline{HOPE ~IT~HELPS}

5 0
3 years ago
What is the following product? Assume x&gt;/=0 ^3sqrt(x^2)*^sqrtx^3
RoseWind [281]

Answer:

x(\sqrt[12]{x^5} )

Step-by-step explanation:

We need to remember 2 rules when doing these:

1. \sqrt[n]{x^a} =x^{\frac{a}{n}}

2. x^a*x^b=x^{ab}

Using these 2 rules, we can simplify the product (steps shown below):

\sqrt[3]{x^2} *\sqrt[4]{x^3} \\=x^{\frac{2}{3}}*x^{\frac{3}{4}}\\=x^{\frac{2}{3}+\frac{3}{4}}\\=x^{\frac{17}{12}}\\=x^{\frac{12}{12}+\frac{5}{12}}\\=x(x^{\frac{5}{12}})\\=x(\sqrt[12]{x^5} )

Rearranging, we see that it is the third choice.

7 0
3 years ago
Expression
Alex787 [66]
The value of the expression is -24

This is because you distribute the -4 to both of the values in the parenthesis like this:
-4(7) + -4(-1)
This can be simplified to:
-28 + 4
Which equals -24
Hope this helps!
4 0
3 years ago
A ship leaves port at noon and travels at a bearing of 214°. The ship’s average rate of speed is 15 miles per hour.
Daniel [21]

Answer: (-31.09 mi, -20.97 mi)

Step-by-step explanation:

If the noon is our initial time, then we have that 2:30 pm is t = 2.5 hours.

Now, the position (x, y) of the ship can be writen as:

(velocity*time*cos(angle), velocity*time*sin(angle))

or:

(15mph*t¨*cos(214°), 15mph*t*sin(214°))

then the position at t = 2.5 hours is:

(15mph*2.5h*cos(214°), 15mph*2.5h*sin(214°))

(-31.09 mi, -20.97 mi)

3 0
4 years ago
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