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MrRa [10]
3 years ago
7

Find the distance between the points (1, 0) and (0, 2).

Mathematics
2 answers:
Nadya [2.5K]3 years ago
8 0
The answer is 2.23606797 approximately
vesna_86 [32]3 years ago
6 0

Answer:

is 2

Step-by-step explanation:

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The cubic polynomial below has a double root at r4 and one root at x = 6 and passes Through the point (2, 36) as shown. Algebrai
qaws [65]

The graph in the question is missing.

Answer:

y  = \frac{-1}{4}( x^3 + 2x^2 - 32x -96)

Step-by-step explanation:

The function is cubic

It has roots as -4, -4 , 6

this means the value of x = -4, -4 , 6 which makes the entire equation zero

so we have solutions as

x+4 = 0

x+4 = 0

x- 6 = 0

on forming a cubic equation using these

(x+4)(x+4)(x-6)

the equation passes through (2,36)

put x = 2

(2+4)(2+4)(2-6) = (6)*(6)*(-4)

which exceeds 36 so we product the equation with -1/4 to get 36

Final equation

y = \frac{-1}{4} (x+4)(x+4)(x-6)

  y  = \frac{-1}{4}( x^3 + 2x^2 - 32x -96)

3 0
2 years ago
Im really bad in math so could anybody help me
Anna71 [15]

Answer:

B , D , A ,C

Step-by-step explanation:

PEMDAS

7 0
3 years ago
Read 2 more answers
HELP PLZ FIRST ANSWER WILL GET BRAINLIEST<br> 5 5/6 times 4 1/4
Nikolay [14]

Hello!

It would be 24.79 but if the answer were to be rounded, it would be 25

Hope this helps! :)

6 0
2 years ago
Read 2 more answers
What is 1 and 1/6 minus 7/12 ?
Damm [24]
1 \frac{1}{6} -  \frac{7}{12} \\ &#10; \\ 1 \frac{1}{6}= \frac{1x6+1}{6}= \frac{7}{6} \\ &#10; \\  \frac{7x2}{6x2}= \frac{14}{12} \\ &#10; \\  \frac{7x1}{12x1}= \frac{7}{12} \\ &#10; \\  \frac{14}{12} -  \frac{7}{12}= \frac{7}{12} \\ &#10; \\ Solution: \frac{7}{12} &#10;
7 0
4 years ago
Help,, i'll mark as brainliest
PtichkaEL [24]

The first question would be distance from the start, as it is steadily going up as time goes on.

The second question would be distance from the end, as it is steadily going down as time goes on.

The third question would be speed, as the speed is staying stable as shown by the straight lines seen within the distance from start/end graphs being linear lines.

7 0
3 years ago
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