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3 years ago

Find the distance between the two points (2, -9) and (-1, 4) round to the nearest tenth .

Mathematics

1 answer:

QveST [7]3 years ago

7 0

Answer:

13.3 units (nearest tenth)

Step-by-step explanation:

$\boxed{Distance \: between \: 2 \: points = \sqrt{ {(x1 - x2)}^{2} + {(y1 - y2)}^{2} } }$

Using the formula above, distance

$= \sqrt{ {[2 - ( - 1)]}^{2} + ( - 9 - 4)^{2} }$

$= \sqrt{ {(2 + 1)}^{2} + ( - 13) {}^{2} }$

$= \sqrt{ {3}^{2} + ( - 13) {}^{2} }$

$= \sqrt{9 + 169}$

$= \sqrt{178}$

= 13.3 units (nearest tenth)

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3 years ago

Which of the following fractions will convert to terminating decimals? A.5/6

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Answer:

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Step-by-step explanation:

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3 years ago

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Find the surface area of the regular pyramid

butalik [34]

Answer:

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3 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

3 years ago

Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arr

bogdanovich [222]

Answer:

The correct answer is

0.5 F, 0.5 f

Step-by-step explanation:

We note the following

Number of colored circles in the box = 14

Number of red circles in the box = 4

Number of purple circles in the box = 6

Number of blue circles in the box = 4

The allele frequency are as follows

Where the frequency is given as

Genotype Frequency Relative frequency

FF = Red 4 4/14 (0.29≈0.3)

Ff = Purple 6 6/14 (0.43≈0.4)

ff = Blue 4 4/14(0.29≈0.3)

Within this population,

We however have 4 FF = 8 F

6 Ff = 6 F + 6 f and

4 ff = 8 f

Total allele = 8+6+6+8 = 28

Relative frequency of F = (8+6)/28 = 14/28 = 0.5

relative frequency of f = (8+6)/28 = 0.5

Therefore the allele frequencies in the palm tree population is

0.5 F, 0.5 f

When in equilibrium we have

However the FF has the product of F×F which is = F² = 0.29 so the frequency of F = √(0.29) = 0.535≈ 0.5

The frequency of Ff is Ff or fF = 0.43 since there is equal number of each allele in Ff we have fF or Ff = Ff = 0.43

Which hives 0.43/2 = F =f ≈ 0.2

and ff = 0.29 so that f = 0.535 ≈ 0.5

Therefore f = F = 0.5 + 0.2 = 0.7

5 0

3 years ago