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2 years ago

Please help, I need the answer soonnn!!!

Mathematics

2 answers:

Vedmedyk [2.9K]2 years ago

5 0

Answer:

line B I hope this help!! good luckkk

Angelina_Jolie [31]2 years ago

3 0

Answer:line b

Step-by-step explanation:

In general, the slope intercept form assumes the formula: y = mx + b.

M is the slope (lesson on slope)

Mnemonic : 'm' means 'move'

B is the y-intercept (lesson on the y-intercept)

Mnemonic: 'b' means where the line begins.

B is where it intercepts so 1 intercepts the y axis and the only line that intercepts the y axis at 1 is line b

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In tuv y is centroid if yw=5b and ty=4b+14 find tw

stellarik [79]

your right Step-by-step explanation:

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3 years ago

What is the value of x ?

Hatshy [7]

Answer:

The value of x will be 2.5

LY will be 9 while KH will be 7.5

Step-by-step explanation:

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6 0

2 years ago

Can someone please!!! Help me

timurjin [86]

The answer to this question is true

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2 years ago

A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15

Salsk061 [2.6K]

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: $Perimeter=2(Length+Width)\\$

Now, Putting values in formula for finding Width w:

$Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8$

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

6 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.

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

=> 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