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nekit [7.7K]
3 years ago
6

Please help with recursive formula for math

Mathematics
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

Hi... U should do like this

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Use the following diagram to solve the problem
nordsb [41]

Answer:

B

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

94° is an exterior angle of the triangle, thus

60 + 2x = 94 ( subtract 60 from both sides )

2x = 34 ( divide both sides by 2 )

x = 17 → B

7 0
3 years ago
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What is the value of q?<br> A.122<br> B.116<br> C.180<br> D.58
Studentka2010 [4]

Step-by-step explanation:

A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. A convex quadrilateral ABCD is cyclic if and only if its opposite angles are supplementary, means

∠q + 58 =180

∠q = 180 -58

∠q = 122 degree

∠p + 90 = 180

∠p = 180 -90

∠p = 90 degree

4 0
3 years ago
Can anyone write this in a notebook for more understanding? Thank you
vfiekz [6]

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Step-by-step explanation:

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3 0
3 years ago
John bought 4 pants that each cost the same amount and a pair of sneakers that cost $65. The items he bought cost a total of $14
Rzqust [24]

Answer:p=$20

Step-by-step explanation:

145-65=80

80/4=20

so each pant costs $20

p=$20

7 0
3 years ago
PLZ HELP ME ☻ <img src="https://tex.z-dn.net/?f=%5C%5B%5Cfrac%7Bxy%7D%7Bx%20%2B%20y%7D%20%3D%201%2C%20%5Cquad%20%5Cfrac%7Bxz%7D%
Yanka [14]

Answer:

x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}

Second equation:

\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}

Third equation:

\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}

Now we use this value for "x" back in equation 1 to solve for "y":

1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}

And finally we solve for the third unknown "z":

\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12

8 0
3 years ago
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