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Luba_88 [7]
3 years ago
14

What value of c make the relation (c,9), (8,7),(6,5) a function

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:10

Step-by-step explanation:

plz leave a like

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Simplified form of 11[(9^2-5^2)/2^2+8]<br> A.224.5<br> B.242<br> C.51.3<br> D.110
aleksandr82 [10.1K]
B. 242
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4 years ago
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Which statement best explains whether y = 3x + 5 is a linear function or a nonlinear function?
katen-ka-za [31]
The answer is letter 4 you can prove this replacing each pair in the function y=3x+5, and it is a straight line because the greatest exponent of this function is 1.
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3 years ago
Job cashed a check for $1,765. The teller gave him nine hundred-dollar bills, seven fifty-dollar bills, nineteen twenty-dollar b
Luden [163]

Answer:

x = 13

Step-by-step explanation:

1765

x = 100 (9) + (7) 50 + (19) 20 + (7) 10 - 1765

x = 900 + 350 + 380 + 70 - 1765

x = 1700 - 1765

x = -65

65/5 = 13

x = 13

5 0
3 years ago
Prove that the sum of the squares of the lengths of the medians of a triangle is three fourths the sum of the squares of the len
aleksandr82 [10.1K]

Answer:

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

Step-by-step explanation:

I've attached an image showing a triangle divided into it's medians.

Now, from the attached image, we see that AB, BC & CD are the lengths of sides of triangle while AD, BE & CF are lengths of the 3 medians of the triangle.

Now, to prove our question, we will use appolonius theorem which states that "the sum of the squares of any of the two sides of a triangle is equal to twice the square on half the third side including twice the square on the median that bisects the third side.

Applying this theorem to the image attached, we have the following;

|AB|² + |AC|² = 2[|AD|² + |BC/2|²]

|AB|² + |BC|² = 2[|BE|² + |AC/2|²]

|AC|² + |BC|² = 2[|CF|² + |AB/2|²]

Adding the 3 equations above gives us;

2|AB|² + 2|BC|² + 2|AC|² = 2|AD|² + |BC|²/2 + 2|BE|² + |AC|²/2 + 2|CF|² + |AB|²/2

Collecting like terms;

(2|AB|² - |AB|²/2) + (2|BC|² - |BC|²/2) + (2|AC|² - |AC|²/2) = 2|AD|² + 2|BE|² + 2|CF|²

Thus gives;

(3/2)[|AB|² + |BC|² + |AC|²] = 2[|AD|² + |BE|² + |CF|²]

Multiply both sides by 2 to give;

3[|AB|² + |BC|² + |AC|²] = 4[|AD|² + |BE|² + |CF|²]

6 0
3 years ago
Use a common denominator to write an equivalent fraction for each fraction. 5/12, 2/9
lapo4ka [179]

Answer: common denominator 108

Step-by-step explanation:

Gcf= 108

9 * 12

45/108 and 24/108

6 0
3 years ago
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