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3 years ago

Suppose the midpoint of (AB) is M= (3,0) if point A has the coordinates of (-1,4) then what are the coordinates of B?

Mathematics

1 answer:

BaLLatris [955]3 years ago

7 0

Answer:

Step-by-step explanation:

Point M being midpoint of AB means:

$x_M-x_A=x_B - x_M\qquad\quad\ \wedge\qquad y_M-y_A=y_B - y_M\\\\ 3-(-1)=x_B-3\qquad\wedge\qquad 0-4=y_B -0\\\\ {}\qquad x_B=7\qquad\qquad\wedge\qquad\qquad \ y_B=-4$

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3 years ago

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Suppose an opaque jar contains 3 red marbles and 10 green marbles. The following exercise refers to the experiment of picking tw

kow [346]

Answer:

$\frac{5}{13}$

Step-by-step explanation:

Given,

Red marbles = 3,

Green marbles = 10,

So, the total marbles = 3 + 10 = 13,

$\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

Since, here replacement is not allowed,

Thus, the probability of getting a green marble and a red marble

= first red and second green + first green second red

$=\frac{3}{13}\times \frac{10}{12}+\frac{10}{13}\times \frac{3}{12}$

$=2\times \frac{3}{13}\times \frac{10}{12}$

$=\frac{30}{78}$

$=\frac{5}{13}$

Note : The probability of getting a green marble first and a red marble second

= $\frac{10}{13}\times \frac{3}{12}$

$=\frac{30}{156}$

$=\frac{5}{26}$

4 0

4 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$