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PilotLPTM [1.2K]
4 years ago
9

⚠️⚠️ Ok I need some help here lol ⚠️⚠️ This is the only one and it’s honestly confusing

Mathematics
2 answers:
nignag [31]4 years ago
7 0
(4,6) :)
As it goes over the X axis, you pretty much just mirror it, over that axis.
VladimirAG [237]4 years ago
6 0

Answer:

the answer is (4,6)

Step-by-step explanation:

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A rectangle on a coordinate plane has vertices L(0, 6), M(8,6), N(8,0), and O(0, 0). What are the dimensions of the
solniwko [45]

Answer:

The whole rectangle is in the 1st dimensions.

For how to do this, you need draw it in a grid chart.

5 0
3 years ago
The area of the shaded region
lbvjy [14]

Answer:

27.74cm^2

Step-by-step explanation:

area of circle: 3 times 3 times pi = 28.26cm^2

area of square: 8 times 7 = 56cm^2

subtract area of circle from square

56-28.26=27.74cm^2

4 0
4 years ago
Max tossed a fair coin 3 times. What is the probability of getting heads in the first two trials and tails in the last trial?  
Whitepunk [10]
When you toss a coin 3 times here is what can happen:
H=heads
T= Tails
HHH                        TTT
<u>HHT</u>                        TTH
HTH                        THT
HTT                        THH
Since 8 things can happen and only one of them is what you want, the probability is 1/8
Another way to think about it is each toss has 1/2 of coming up as you want it to.
Three tosses are 1/2 x 1/2 x 1/2 which is 1/8
                   
6 0
4 years ago
Read 2 more answers
Dan drew the line of best fit on the scatter plot shown below: A graph is shown with scale along x axis from 0 to 10 at incremen
Anit [1.1K]
     0, 3
-   10, 15
= -10, -12
therefore, the slope is 6/5, and the intercept (c) is as supplied, 3.
the equation, y=mx+c or y = a + bx, can be applied here where m or b = 6/5, and a or c = 3.

therefore the equation is y=6/5x+3.

To test this, you can put in y = 10(6/5)+3, which spits out y = 15. This way we know it *should* work.

3 0
3 years ago
Use logarithmic differentiation to find dy/dx: y=((x^3)(2x+3)^1/2) / (x-2)^2
sukhopar [10]
y=\frac { { { x }^{ 3 }\left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } }{ { \left( x-2 \right)  }^{ 2 } }

\\ \\ { \left( x-2 \right)  }^{ 2 }\cdot y={ x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 }\cdot y \right)  } =\ln { \left( { x }^{ 3 }{ \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ \ln { \left( { \left( x-2 \right)  }^{ 2 } \right)  } +\ln { y } =\ln { \left( { x }^{ 3 } \right)  } +\ln { \left( { \left( 2x+3 \right)  }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ 2\ln { \left( x-2 \right)  } +\ln { y } =3\ln { x } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \ln { y } =3\ln { x } -2\ln { \left( x-2 \right)  } +\frac { 1 }{ 2 } \ln { \left( 2x+3 \right)  }

\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }

\\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =y\left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)

\\ \\ \frac { dy }{ dx } =\frac { { x }^{ 3 }\sqrt { 2x+3 }  }{ { \left( x-2 \right)  }^{ 2 } } \cdot \left( \frac { 3 }{ x } -\frac { 2 }{ x-2 } +\frac { 1 }{ 2x+3 }  \right)
4 0
3 years ago
Read 2 more answers
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