All categories

3 years ago

5.6 x N= 2.24 N=? Please help

Mathematics

2 answers:

Firdavs [7]3 years ago

7 0

Answer:

N= 0.4

Step-by-step explanation:

5.6 x N= 2.24

Divide both sides by 5.6

5.6 and 5.6 cancel out

2.24/5.6= 0.4

Check your work

5.6 x 0.4 =2.24

0.4 is correct

Hope this helps ^-^

dlinn [17]3 years ago

3 0

Answer:

N = -3.36

Step-by-step explanation:

5.6 x N= 2.24

-5.6 -5.6

N= -3.36

You might be interested in

Quotient z on 7 Decreases on 1 Decreases 7 ? ناتج قسمة Z على 7 يقل عن 1 بمقدار 7 ؟

Crank

Answer:

pls I don't understand can u pls expatriate

7 0

3 years ago

Which best describes the dna of a cloned animal

pychu [463]

I think it has the same DNA as its parent

3 0

3 years ago

Read 2 more answers

Locate the absolute extrema of the function on the closed interval: y = 3x^(2/3) - 2x, [-1, 1]

Alecsey [184]

To get the extrema, derive the function.
You get y' = 2x^-1/3 - 2.
Set this equal to zero, and you get x=0 as the location of a critical point.
Since you are on a closed interval [-1, 1], those points can also have an extrema.
Your min is right, but the max isn't at (1,1). At x=-1, you get y=5 (y = 3(-1)^2/3 -2(-1); (-1)^2/3 = 1, not -1).
Thus, the maximum is at (-1, 5).

7 0

3 years ago

The function below models the voltage, in volts, of a certain alternating current after x seconds, where A and b are positive co

Katarina [22]

First of all, we can just ignore A, it has no effect but to vertically stretch our cosine.

If it was only $f(x)=cosx$ , the function would be invertible as long as it's confined between $0$ and $\pi$ . Now, the argument of our cosine is not $x$ but $bx$ . It means that it won't stop at $\pi$ , but at $\frac{\pi}{b}$ .

Another way to think about it, "what should i replace x with so I get $\pi$ inside the cosine?"

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.

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

=> 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