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mihalych1998 [28]
3 years ago
6

For a class, Mrs. Hawk brought

Mathematics
2 answers:
Mekhanik [1.2K]3 years ago
6 0

Answer:

14

Step-by-step explanation:

7x2=14

tangare [24]3 years ago
3 0
7C2 = 7!/[2!(7-2)!] = 7!/(2x5!) = 7x6/2 = 21
Therefore there are 21 possible combinations.
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Regina left the coffee shop traveling 9 mph. Then, 3 hours later, Janai left traveling the same direction at
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Answer:

<em>suwi6q74diyer7ityw7reuDjuss36audlsgo74wridoyaurslhjsgjfaufalakateUd</em>

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will3urafg-"7'-"/-()4<u> </u><u>mhl374</u><u>y</u><u>3</u><u>l</u><u>u</u><u>s</u><u>d</u><u>u</u><u>s</u><u>i</u><u>t</u><u>z</u><u>l</u><u>a</u><u>7</u><u>w</u><u>h</u><u>f</u><u>z</u><u>s</u><u>p</u><u>r</u><u>X</u><u>j</u><u>a</u><u>u</u><u>e</u><u>g</u><u>k</u><u>z</u><u>h</u><u>r</u><u>o</u><u>y</u>

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17) please help with this question!!!<br> BraiNLIEST AND 20 POINTS :)
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A

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4 0
3 years ago
The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

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

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

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

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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Answer:

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8 0
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