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dimulka [17.4K]
3 years ago
6

Can somebody help me I don’t know which one is the answer!!!

Mathematics
1 answer:
emmasim [6.3K]3 years ago
6 0
The answer is C. Steps shown below.

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The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi
slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

y(x)=\frac{1}{\sqrt{C+4x^2}}

Now for the initial condition the value of C is calculated as

y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16

So the solution is given as

y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}

Simplifying the equation as

y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}

So the correct option is 4

8 0
3 years ago
Read 2 more answers
How many and what type of solutions does 7x^2−4x+3 have? 1 rational solution 2 rational solutions 2 irrational solutions 2 nonre
Gre4nikov [31]

Answer:

2 non real solutions.

Step-by-step explanation:

We need to use discriminant,

for ax²+bx+c=0

The discriminat is b²-4ac

If the discriminant is,

→ less than 0, then 0 real solutions

→ equal to 0, then 1 real solutions

→ more than 0, then 2 real solutions

Given that,

7x²−4x+3=0

a=7, b=-4, and c=3

→ (-4)²-4(7)(3)

→ 16-84

→ -68

You can see this is less than 0, then non real solutions. [2 nonreal solutions]

7 0
3 years ago
Read 2 more answers
Which of the binomials below is a factor or this trinomial x-2 +5x+6
lutik1710 [3]
Firstly, for it to be a trinomial the expression would have to be x^2+5x+6 to factor this we would simply have to find 2 numbers that add up to 5 and multiply to 6. 3+2=5 and 3•2=6 so (x+3)(x+2) would be the answer
5 0
3 years ago
What is the tan invers of 3i/-1-i​
postnew [5]

<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> )

<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> ) × (-1 + <em>i</em> ) / (-1 + <em>i</em> )

<em>z</em> = (3<em>i</em> × (-1 + <em>i</em> )) / ((-1)² - <em>i</em> ²)

<em>z</em> = (-3<em>i</em> + 3<em>i</em> ²) / ((-1)² - <em>i</em> ²)

<em>z</em> = (-3 - 3<em>i </em>) / (1 - (-1))

<em>z</em> = (-3 - 3<em>i </em>) / 2

Note that this number lies in the third quadrant of the complex plane, where both Re(<em>z</em>) and Im(<em>z</em>) are negative. But arctan only returns angles between -<em>π</em>/2 and <em>π</em>/2. So we have

arg(<em>z</em>) = arctan((-3/2)/(-3/2)) - <em>π</em>

arg(<em>z</em>) = arctan(1) - <em>π</em>

arg(<em>z</em>) = <em>π</em>/4 - <em>π</em>

arg(<em>z</em>) = -3<em>π</em>/4

where I'm taking arg(<em>z</em>) to have a range of -<em>π</em> < arg(<em>z</em>) ≤ <em>π</em>.

6 0
2 years ago
Rachel saved money to buy art supplies. She used 13 of her savings to buy brushes. She used 35 of her savings to buy paint. What
vekshin1

Given:

Rachel used \dfrac{1}{3} of her savings to buy brushes.

She used \dfrac{3}{5} of her savings to buy paint.

To find:

The fraction for remaining savings.

Solution:

Fraction that Rachel used of her savings to buy brushes and paint is

\text{Fraction for used amount}=\dfrac{1}{3}+\dfrac{3}{5}

                                    =\dfrac{5+9}{15}

                                    =\dfrac{14}{15}

Now,

\text{Fraction for remaining savings}=1-\text{Fraction for used amount}

\text{Fraction for remaining savings}=1-\dfrac{14}{15}

\text{Fraction for remaining savings}=\dfrac{1}{15}

Therefore, \dfrac{1}{15} of her savings is remaining.

6 0
3 years ago
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