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professor190 [17]
1 year ago
15

Select the correct answer. A linear function graph of the x-axis and y-axis has a diagonal line that passes the x-axis at (minus

0.5, 0), and the y-axis at (0, 1) In the function above, the slope will be multiplied by , and the y-value of the y-intercept will be increased by 3 units. Which of the following graphs best represents the new function?

Mathematics
1 answer:
BARSIC [14]1 year ago
5 0

Answer: Graph Y

Step-by-step explanation:

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A 3cm×10cm rectangle sits inside a circle with radius of 12. Find the area of the shaded region
kolezko [41]

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422.4cm²

Step-by-step explanation:

area of circle is pi x 12² which is 144 pi

144pi subtract 30 is 422.4

6 0
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Cube M has an edge length of
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4

Step-by-step explanation:

6*(8*2)^2/(6*8^2)=2^2=4

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How do I do the power rule I don't understand
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You multiply the number by itself however many times it says for example 2 to the second power 2x2 is 4
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the dimensions of a triangular prism are shown in the diagram.What is the volume of the triangular prism in cubic centimeters
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Step-by-step explanation:

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= 1632

Step-by-step explanation:

7 0
2 years ago
Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u
Tamiku [17]

Answer:

Step-by-step explanation:

15. \frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx}   \\

16.  \frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }

=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2}  \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\

=> The clause is correct

17. Do the same (16)

18. \frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx

19.

\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\

=> The clause is correct

20. Do the same (18)

21. Left = \frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA}  } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}

Right = cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA}   \\

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0
3 years ago
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