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bezimeni [28]
3 years ago
10

PLZ HELP ILL MARK BRAINLIEST to the best

Mathematics
2 answers:
larisa [96]3 years ago
8 0
I believe it is a Positive Linier association
Tatiana [17]3 years ago
7 0
Positive linear association
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A + (−5) = −12<br><br> What is a
kherson [118]
A + -5 = -12

A - 5 = -12

A = -7.

This is the solution.
7 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!
Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

  • if "a" is positive, then opens UP
  • if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

  • the Axis Of Symmetry (x-value) is: x = \frac{-b}{2a}  
  • y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

  • a = +1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-11}{2(1)}  = -\frac{11}{2}
  • y = (-\frac{11}{2})^{2} + 11(-\frac{11}{2} ) + 24 = -\frac{25}{4}
  • vertex (-\frac{11}{2}, -\frac{25}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

  • a = -1 so the parabola opens DOWN
  • x = \frac{-b}{2a} = \frac{-(-6)}{2(-1)}  = \frac{6}{-2} = -3
  • y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1
  • vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-(-2)}{2(1)}  = \frac{2}{2} = 1
  • y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2
  • vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). <em>There are no x-intercepts</em>

******************************************************************************************

4) y = x² + 4x + 4

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-4}{2(1)}  = -2
  • y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0
  • vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). <em>The x-intercept is the vertex.</em>

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

  • a = +3 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-21}{2(3)}  = -\frac{7}{2}
  • y = 3(-\frac{7}{2})^{2} + 21(-\frac{7}{2} ) + 30 = -\frac{27}{4}
  • vertex (-\frac{7}{2}, -\frac{27}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

*******************************************************************************************

4 0
3 years ago
Read 2 more answers
Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx
uranmaximum [27]
Given the differential equation

5t^3 \frac{dx}{dt} +2x-2=0

The solution is as follows:

5t^3 \frac{dx}{dt} +2x-2=0 \\  \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\  \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\  \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\  \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\  \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\  \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\  \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\  \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1
3 0
3 years ago
The perimeter of a rectangular store receipt is ten and five-sixths inches. If three of the sides total eight and one-half inche
serg [7]

Answer:

The measure of the fourth side is \frac{7}{3} .

Step-by-step explanation:

Given as :

The Perimeter of rectangular store =  ( 10 + \frac{5}{6} )  inches

The measure of three sides of rectangle =  ( 8 + \frac{1}{2} )  inches

So, The measure of three sides = length + breadth + length

I,.e   2 L + B =  ( 8 + \frac{1}{2} )  inches      .....1

∵ perimeter =  ( 10 + \frac{5}{6} )  inches

So ,  2 L + 2 B = ( 10 + \frac{5}{6} )  inches

Or,  2 L + B + B = ( 10 + \frac{5}{6} )  inches

Or, ( 8 + \frac{1}{2} ) + B =  ( 10 + \frac{5}{6} )

or, B =  ( 10 + \frac{5}{6} ) -  ( 8 + \frac{1}{2} )

Or, B = 10 - 8 +  (\frac{5}{6} ) -  \frac{1}{2}

or, B = 2 +  \frac{5-3}{6}

or, B = 2 +  \frac{1}{3}

∴   B =  \frac{6 + 1}{3}  =  \frac{7}{3}

Hence The measure of the fourth side is \frac{7}{3} . Answer

7 0
3 years ago
QUICK HELP!!!!!!!!!!!!
Trava [24]

Answer:

50%

Step-by-step explanation:

8+3+4=15

15/3=5

5x10=50

3 0
3 years ago
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