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vlabodo [156]
2 years ago
11

Double g, then subtract f from the result

Mathematics
1 answer:
Maurinko [17]2 years ago
8 0

Answer:

2g - f

Step-by-step explanation:

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How many solutions does this equation have?<br><br> a + 3 + 2a = -1 + 3a + 4
vekshin1
-8a+4 im pretty sure 

4 0
2 years ago
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Linda knitted 1/6 of a sweater on Friday and 2/9 of the sweater on Saturday. On Sunday she knitted 1/2 of the remaining sweater.
sladkih [1.3K]
This is pretty simple. All you have to do is make the first and second fraction share a denominator by multiplying them by each other.

6 <span>• 9 = 54

Then multiply each numerator by the opposing denominator.

1 </span><span>• 9 = 9

2 </span><span>• 6 = 12

Here are the new fractions:

9/54

12/54

Now add the 9 and 12 together.

9 + 12 = 21

The complete fraction:

21/54

Subtract 21 from 54 so you can get the remainder of the sweater.

54 - 21 = 33

This is the remainder fraction:

33/54

Can you simplify this? Yes, of course! They can both be divided by 3!

11/18

That is the remainder of the sweater. But you still have to divide it in half! After all, Linda only knitted half of the remaining sweater. Dividing it in half can be done just by multiplying the denominator by 2.

11/36

That should be your answer! Apologies if I got something wrong.</span>
7 0
3 years ago
Is there a mistake in this equation? if not please let me know!
Juli2301 [7.4K]

Answer:

no mistakes

Step-by-step explanation:

4 * 3^x = 324

Divide each side by 4

4/4 * 3^x = 324/4

3^x = 81

Rewriting 81 as a power of 3

3^x = 3^4

Since the bases are the same, the powers are the same

x=4

5 0
3 years ago
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Triangle \triangle A'B'C'△A
ioda

Answer:

the awnser is A

Step-by-step explanation:

7 0
2 years ago
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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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