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padilas [110]
3 years ago
11

Mjnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Mathematics
1 answer:
Radda [10]3 years ago
7 0

I don't know what this is supposed to mean, but thanks for the free points! ;)

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What will the answer be
PSYCHO15rus [73]
Since 1 is your initial value you plot a point at (0,1) just like you did , but since your slope (rise/run) is 3, you will go up 3 and over 1, so you will need to graph the point (1,4) instead of (1,-2). Your line will be going up , since a graph is read left to right, you have a positive slope so your line will go up. Hope I helped !
8 0
2 years ago
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The graph of y = x^2 -3 is translated in 4 units to the left of the graph to give the graph A
Vedmedyk [2.9K]

Answer:

a) b = 8, c = 13

b) The equation of graph B is y = -x² + 3

Step-by-step explanation:

* Let us talk about the transformation

  • If the function f(x) reflected across the x-axis, then the new  function g(x) = - f(x)
  • If the function f(x) reflected across the y-axis, then the new  function g(x) = f(-x)
  • If the function f(x) translated horizontally to the right  by h units, then the new function g(x) = f(x - h)
  • If the function f(x) translated horizontally to the left  by h units, then the new function g(x) = f(x + h)

In the given question

∵ y = x² - 3

∵ The graph is translated 4 units to the left

→ That means substitute x by x + 4 as 4th rule above

∴ y = (x + 4)² - 3

→ Solve the bracket to put it in the form of y = ax² + bx + c

∵ (x + 4)² = (x + 4)(x + 4) = (x)(x) + (x)(4) + (4)(x) + (4)(4)

∴ (x + 4)² = x² + 4x + 4x + 16

→ Add the like terms

∴ (x + 4)² = x² + 8x + 16

→ Substitute it in the y above

∴ y = x² + 8x + 16 - 3

→ Add the like terms

∴ y = x² + 8x + 13

∴ b = 8 and c = 13

a) b = 8, c = 13

∵ The graph A is reflected in the x-axis

→ That means y will change to -y as 1st rule above

∴ -y = (x² - 3)

→ Multiply both sides by -1 to make y positive

∴ y = -(x² - 3)

→ Multiply the bracket by the negative sign

∴ y = -x² + 3

b) The equation of graph B is y = -x² + 3

4 0
3 years ago
The velocity of a particle moving back and forth on a line is vequalsStartFraction ds Over dt EndFraction equals6 sine (2 t )Sta
max2010maxim [7]

Answer:

s = 6 m

Step-by-step explanation:

The value of the velocity v is given as:

v = \frac{ds}{dt} = 6 sin(2t) m/s

To find s, we have to integrate and apply the initial values of s = o when t = 0:

\frac{ds}{dt} = 6 sin(2t)\\\\\int\limits^s_0 {ds} = \int\limits^t_0 {6sin(2t)} \, dt\\\\s|^s_o = -3cos(2t)|^t_o\\\\s - 0 = -3cos(2t) -(-3cos(0))\\\\s = -3cos(2t) + 3(1)\\\\s = -3cos(2t) + 3

When t = π/2, s will be:

s = -3cos(2 * π/2) + 3

s = -3cos(π) + 3

s = -3(-1) + 3

s = 3 + 3

s = 6 m

4 0
3 years ago
What is the answer I need help!
g100num [7]

Answer:

B

Step-by-step explanation:

y³=64

y=(64)^{\frac{1}{3} } =\sqrt[3]{64}

5 0
2 years ago
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If the m∠SRW = 85°, what are the measures of ∠VRU and ∠URW?<br><br> m∠VRU = °<br><br> m∠URW = °
alina1380 [7]

Answer:

see explanation

Step-by-step explanation:

∠TRV = ∠SRW ( vertical angles ), hence

2x + 15 = 85 ( subtract 15 from both sides )

2x = 70 ( divide both sides by 2 )

x = 35

Hence

∠VRU = 2x + 10 = ( 2 × 35) + 10 = 70 + 10 = 80°

and

∠URW + ∠VRU + ∠SRW = 180 ( sum of angles in a straight line ), thus

∠URW + 80 + 85 = 180

∠URW + 165 = 180 ( subtract 165 from both sides )

∠URW = 15°

5 0
2 years ago
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