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zmey [24]
3 years ago
6

Simplify square root of -40

Mathematics
2 answers:
faltersainse [42]3 years ago
5 0

if you know complex numbers:


sqrt (-40)

sqrt(-1) * sqrt(40)

sqrt(-1) * sqrt(4) * sqrt(10)

i * 2 * sqrt(10)

2i *sqrt(10)


If you do not know complex numbers:

it cannot be simplified because it is negative

no real solution

s2008m [1.1K]3 years ago
5 0

Answer: 2<em>i</em> \sqrt{10}

Step-by-step explanation: To simplify the square root of -40, think of the square root of -40 as the square root of -1 × the square root of 40.

The square root of -1 can be represented by the imaginary number <em>i</em>. So we have <em>i</em> × the square root of 40 or <em>i</em> \sqrt{40}.

Next, root 40 breaks down breaks down to 2 root 10. We can see this by setting up a factor tree for 40. 40 is 2 x 20, 20 is 2 x 10, and 10 is 2 x 5. We have a pair of 2's in our factor tree and we have 2 x 5 at the bottom.

So we have <em>i</em> x 2 root 10 or 2<em>i</em> \sqrt{10}.

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Write4×4×4×4 in index​
JulijaS [17]

Answer:

4^4

Step-by-step explanation:

the answer is 4 to the power of 4

5 0
2 years ago
23. The letters of the alphabet are written on
Alla [95]

Answer:

21/676

Step-by-step explanation:

The total number of letters of the alphabet will be the total outcome which is 26 alphabets.

If consonant numbers are drawn out, the probability of drawing a consonant will be total consonants/total alphabets

Total consonants in the alphabet is 21.

Probability of drawing consonants = 21/26.

Since all the consonants are replaced before drawing a Z, our total outcome will not change i.e it will still be 26.

Probability of drawing a letter Z will be 1/26.

Therefore, the probability of drawing a consonant,replacing it, and then drawing a Z will be;

21/26×1/26

= 21/676

5 0
3 years ago
What is the equation of the line that passes through (1, 2) and is parallel to the line whose equation is 4x + y + 1 = 0?
Bezzdna [24]

Answer:

The answer is

<h2>4x + y - 6 = 0</h2>

Step-by-step explanation:

Equation of a line is y = mx + c

where m is the slope

c is the y intercept

4x + y + 1 = 0

y = - 4x - 1

Comparing with the above formula

Slope / m = - 4

Since the lines are parallel their slope are also the same

That's

Slope of the parallel line is also - 4

Equation of the line using point ( 1 , 2) is

y - 2 = -4(x - 1)

y - 2 = - 4x + 4

4x + y - 2 - 4

We have the final answer as

<h3>4x + y - 6 = 0</h3>

Hope this helps you

7 0
3 years ago
Shannon graphed the system of equations.
ehidna [41]
All you need to do is plug -5 into the second equation and you see it is near (-5, -8). When plugged into the top, you get (-5, -27/4) which comes out to ABOUT -6.75 for the Y value. The closest is actually a tie. The first option is .8 from the first and .45 from the second leading in a total distance of 1.25. The second, which is the fellow answer, is 1.2 from the first and .05 from the second, leading to 1.25 away.

The third, which is next closest is 1.8 from the first and .55 from the second leading to a distance of over 2 from the optimal, so only the first two are answers.
3 0
3 years ago
Read 2 more answers
Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!
julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

y_1 is not a solution of the differential equation

 y_2 is not a solution of the differential equation

y_3 is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is y'' + y' = cos2x

Let consider the first equation to substitute

y_1  = cosx  +sinx

y_1'  = -sinx  +cosx

y_1''  = -cosx -sinx

So

y_1'' - y_1'  = -cosx -sinx -sinx  +cosx

y_1'' + y_1'  = -2sinx

So

-2sinx \ne  cos2x

This means that y_1 is not a solution of the differential equation

Let consider the second equation to substitute

y_2 =  cos2x

y_2' =  -2sin2x

y_2'' =  -4cos2x

So

y_2'' + y_2'  = -4cos2x-2sin2x

So

-4cos2x-2sin2x \ne  cos2x

This means that y_2 is not a solution of the differential equation

Let consider the third equation to substitute

y_3 =  sin 2x

   y_3' =  2cos 2x

    y_3'' =  -4sin2x

So

y_3'' + y_3'  = -4sin2x  - 2cos2x

So

-4sin2x  - 2cos2x \ne  cos2x

This means  that  y_3 is not a solution of the differential equation

6 0
2 years ago
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