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3 years ago

HELLLPPPPPPPP ASAP WILL GIVE BRAINLIESTTTTTTT ALONG WITH 98 POINTSSSS!!! Make a pie chart.

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

Substitute for x then you will get the answer

weeeeeb [17]3 years ago

6 0

this is real easy to do just substitue for X you can do this man mark me as brainlest please!!!!!!!

You might be interested in

Which relation is a function?

zimovet [89]

Answer:

Step-by-step explanation:

A relation is a set of related points. A function is a set of related points where no inputs repeat.

A. {(1, 2), (2, 3), (3, 2), (2, 1)}

This repeats (2,3) and (2,1). Not a function.

B. {(4, 2), (3, 3), (2, 4), (3, 2)}

This repeats (3,3) and (3,2). Not a function.

C. {(1, −1), (−2, 2), (−1, 2), (1, −2)}

This repeats (1,-1) and (1,-2). Not a function.

D. {(1, 4), (2, 3), (3, 2), (4, 1)}

This doe NOT repeat. Function.

6 0

3 years ago

Amelie found the solutions of the equation x2 − 1x = 42 to be 6 and −7. Explain why this answer is incorrect. Then, find the cor

cupoosta [38]

I dont know how to solve it but since the answer shows as 6 and -7, if you plug it in as x values, one tiem as 6 and second time as -7 it doesnt equal 42

7 0

3 years ago

How do I solve this question?

Montano1993 [528]

Answer:

p = $\frac{3}{2}$ , q = 9

Step-by-step explanation:

4x² + 12x ( factor out 4 from each term )

= 4(x² + 3x)

Using the method of completing the square

add/subtract ( half the coefficient of the x- term)² to x² + 3x

= 4(x² + 2( $\frac{3}{2}$ )x + $\frac{9}{4}$ - $\frac{9}{4}$ )

= 4(x + $\frac{3}{2}$ )² - 4 × $\frac{9}{4}$

= 4(x + $\frac{3}{2}$ )² - 9

4 0

3 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.

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