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sergeinik [125]
3 years ago
13

Find the amount of time that has elapsed.Assumed that the given below are of the same day.write your solution.

Mathematics
1 answer:
belka [17]3 years ago
8 0
1: ----> 2:00
2: ----> 4:45
3: -----> 2:15
4: -----> 4:41
5: -----> 16:15


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If angle 1 is 125 and angle 3 is 3x -1 what is x
Amiraneli [1.4K]

Answer: X means you have to times

Step-by-step explanation: So you will have to do 3x3x3

8 0
2 years ago
Can someone solve and explain how to do these absolute equations for 40 points? I'll give brainliest to the best correct answer.
irga5000 [103]

Answer:

Explanation Below

Step-by-step explanation:

1.    -4 + | x - 1 | = 3

= -4 + | x - 1 | + 4 = 3 + 4

= | x - 1 | = 7

= x - 1 = -7     or    x - 1 = 7

= x = -6 or x = 8

2.   6 + 4 | 2x + 6| = 14

 = 6 + 4 | 2x + 6 | - 6 = 14 - 6

 = 4 | 2x + 6 | = 8

 = 4 | 2x + 6 | / 4 = 8/4

 = 2x + 6 = -2          or         2x + 6 = 2

 = x = -4    or     x = -2

3.  -10 + | -6 - x | = 1

  = -10 + | -6 + x | + 10 = 1 + 10

  = | -6 - x | = 11

  =  -6 - x = -11     or   -6 - x = 11

  = x = -17        or       x = 5

4.    | 7 - 2x | + 4 = 5

   = | 7 - 2x | + 4 - 4 = 5 - 4

   = | 7 - 2x | = 1

   = 7 - 2x = -1            or             7 - 2x = 1

   = x = 3       or   x = 4

Here are the Steps:

SOLVING EQUATIONS CONTAINING ABSOLUTE VALUE(S)

Step 1: Isolate the absolute value expression.

Step2: Set the quantity inside the absolute value notation equal to + and - the quantity on the other side of the equation.

Step 3: Solve for the unknown in both equations.

Step 4: Check your answer analytically or graphically.

Hopes this Helps :D

Mark me Brainliest : )

4 0
2 years ago
The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet
vladimir1956 [14]
\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}

\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
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8 0
3 years ago
Read 2 more answers
What is the y-intercept of D=-30t + 300?
Harlamova29_29 [7]

Answer:

150

Step-by-step explanation:

5 0
2 years ago
Which of the following points is the solution of the quadratic equation:3x^2-5x-2=0​
Maslowich

Answer:

x1 = 2

x2 = -1/3

Step-by-step explanation:

3x^2-5x-2=0

D=25-4*(-2)*3 = 25 + 24 = 49 --> sqrt 49 = 7

x1 = 5 + 7 / 2*3 =   2

x2 = 5 - 7 / 2*3 =  -1/3

7 0
2 years ago
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