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Yakvenalex [24]
3 years ago
8

Geometry - 1.6 - HOMEWORK

Mathematics
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

10/-3

Step-by-step explanation:

use the equation y2-y1/x2-x1

put it in

5+5/2-5

10/-3

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How many ways can David pick four of the first twelve positive integers such that no two of the numbers he picks are consecutive
vekshin1

Answer:

70 times

Step-by-step explanation:

5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 The smallest integer in our sequence is 2n = 2*17 = 34 Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70

6 0
3 years ago
Find the 6th term of the expansion of (2p - 3q)11. a. -7,185,024p4q7 c. -7,185p4q7 b. -7,185,024p6q5 d. -7,185p6q5
bezimeni [28]
\bf (2p-3q)^{11}\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2p)^{11}(-3q)^0\\
2&+11&(2p)^{10}(-3q)^1\\
3&+55&(2p)^9(-3q)^2\\
4&+165&(2p)^8(-3q)^3\\
5&+330&(2p)^7(-3q)^4\\
6&+462&(2p)^6(-3q)^5
\end{array}

the coefficient for the first term is 1, the next is 11 and so on... now, notice, the elements of the binomial, the 1st element starts off with 11, and every term it goes down by 1, the 2nd element starts off at 0, and goes up by 1 in each term.

now, to get the next coefficient, you simply, "get the product of the current coefficient and the exponent of the 1st element, and divide that by the exponent of the 2nd element in the next term".

for example, how did we get 165 for the 4th term.... well  (55*9)/3

how did we get 462 for the 6th term? well (330*7)/5.

and then you can just expand it from there.
8 0
3 years ago
Read 2 more answers
Vector B has x, y, and z components of 3, 7.2,
Temka [501]

Answer:

\approx 9.0\:\mathrm{units}

Step-by-step explanation:

The magnitude of 3D vector \vec{v} can be given by the following:

| \overline{v}| =\sqrt{{v_x}^2+{v_y}^2+{v_z}^2}

Plugging in given values, we have:

| \overline{v}|=\sqrt{3^2+7.2^2+4.5^2},\\| \overline{v}| \approx \fbox{$9.0\:\mathrm{units}$}.

6 0
2 years ago
Can somebody help plz​
Evgen [1.6K]

Answer:

(3,2)

Step-by-step explanation:

the other three ordered pairs do not lie in the shaded region

6 0
3 years ago
Read 2 more answers
This is due in 20 minutes please show work
Sav [38]

Answer:

p,*,fggfgffgggggggg

70

5 0
3 years ago
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