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3 years ago

Three times a number, plus 6.4, is 13.9

Mathematics

1 answer:

yawa3891 [41]3 years ago

5 0

The unknown number equals to 2.5

3x+6.4=13.9
Subtract 6.4 from both sides
3x=7.5
Divide 3 on both sides
X=2.5

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What is the sum of StartRoot negative 2 EndRoot and StartRoot negative 18 EndRoot? 4 StartRoot 2 EndRoot 4 StartRoot 2 EndRoot i

Sveta_85 [38]

Adding $\sqrt{-2}\,\,and\,\,\sqrt{-18}$ we get $4\sqrt{2i}$

Option C 4 StartRoot 2 EndRoot i is correct.

Step-by-step explanation:

We need to find the sum of: $\sqrt{-2}\,\,and\,\,\sqrt{-18}$

Finding sum:

$\sqrt{-2}+\sqrt{-18}$

$\sqrt{-2}$ can be written as $\sqrt{2}i$

$\sqrt{-18}$ can be written as $\sqrt{18}i$

We know 18 = 2x3x3 = 2x3^2 so,

$\sqrt{2i}+\sqrt{2\times 3^2}i$

$\sqrt{2i}+3\sqrt{2}i$

Adding like terms:

$4\sqrt{2i}$

So, adding $\sqrt{-2}\,\,and\,\,\sqrt{-18}$ we get $4\sqrt{2i}$

Option C 4 StartRoot 2 EndRoot i is correct.

Keywords: Square root expressions

Learn more about Square root expressions at:

#learnwithBrainly

6 0

3 years ago

Read 2 more answers

ANSWER THE IMAGINE ABOVE, TYYSM<3

jeka94

It is definitely a!!!!

8 0

2 years ago

Pleaseeeee ı dont find the answer

mote1985 [20]

Answer: $y(x) = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

==========================================================

Explanation:

The given differential equation (DE) is

$y'-\frac{7}{x}y = \frac{y^3}{x^8}\\\\$

Which is the same as

$y'-\frac{7}{x}y = \frac{1}{x^8}y^3\\\\$

This 2nd DE is in the form $y' + P(x)y = Q(x)y^n$

where

$P(x) = -\frac{7}{x}\\\\Q(x) = \frac{1}{x^8}\\\\n = 3$

As the instructions state, we'll use the substitution $u = y^{1-n}$

We specifically use $u = y^{1-n} = y^{1-3} = y^{-2}$

-----------------

After making the substitution, we'll end up with this form

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\$

Plugging in the items mentioned, we get:

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\\frac{du}{dx} + (1-3)*\frac{-7}{x}u = (1-3)\frac{1}{x^8}\\\\\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$

We can see that we have a new P(x) and Q(x)

$P(x) = \frac{14}{x}\\\\Q(x) = -\frac{2}{x^8}$

-------------------

To solve the linear DE $\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$ , we'll need the integrating factor which I'll call m

$m(x) = e^{\int P(x) dx} = e^{\int \frac{14}{x}dx} = e^{14\ln(x)}$

$m(x) = e^{\ln(x^{14})} = x^{14}$

We will multiply both sides of the linear DE by this m(x) integrating factor to help with further integration down the road.

$\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\m(x)*\left(\frac{du}{dx} + \frac{14}{x}u\right) = m(x)*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + x^{14}*\frac{14}{x}u = x^{14}*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + 14x^{13}*u = -2x^6\\\\\left(x^{14}*u\right)' = -2x^6\\\\$

It might help to think of the product rule being done in reverse.

Now we can integrate both sides to solve for u

$\left(x^{14}*u\right)' = -2x^6\\\\\displaystyle \int\left(x^{14}*u\right)'dx = \int -2x^6 dx\\\\\displaystyle x^{14}*u = \frac{-2x^7}{7}+C\\\\\displaystyle u = x^{-14}*\left(\frac{-2x^7}{7}+C\right)\\\\\displaystyle u = x^{-14}*\frac{-2x^7}{7}+Cx^{-14}\\\\\displaystyle u = \frac{-2x^{-7}}{7}+Cx^{-14}\\\\$

$u = \frac{-2}{7x^7} + \frac{C}{x^{14}}\\\\u = \frac{-2}{7x^7}*\frac{x^7}{x^7} + \frac{C}{x^{14}}*\frac{7}{7}\\\\u = \frac{-2x^7}{7x^{14}} + \frac{7C}{7x^{14}}\\\\u = \frac{-2x^7+7C}{7x^{14}}\\\\$

Unfortunately, this isn't the last step. We still need to find y.

Recall that we found $u = y^{-2}\\\\$

So,

$u = \frac{-2x^7+7C}{7x^{14}}\\\\y^{-2} = \frac{-2x^7+7C}{7x^{14}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7C}$

We're told that y(1) = 1. This means plugging x = 1 leads to the output y = 1. So the RHS of the last equation should lead to 1. We'll plug x = 1 into that RHS, set the result equal to 1 and solve for C

$\frac{7*1^{14}}{-2*1^7+7C} = 1\\\\\frac{7}{-2+7C} = 1\\\\7 = -2+7C\\\\7+2 = 7C\\\\7C = 9\\\\C = \frac{9}{7}$

So,

$y^{2} = \frac{7x^{14}}{-2x^7+7C}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7*\frac{9}{7}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+9}\\\\y = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

We go with the positive version of the root because y(1) is positive, which must mean y(x) is positive for all x in the domain.

3 0

2 years ago

Carrie found the difference 11/12 -- 2/3. she said * i found a common denominator 12,then subtracted the numerators to get 9/12*

nikitadnepr [17]

Answer:

She isn't correct because after you find the common denominator, the numerator needs to be changed as well, so when she multiplied the denominator of 2/3 by 4 to get 12, she also had to multiply the 2 by 4, to get 8/12, and then subtracted, because otherwise, the new fraction isn't equivalent to the original fraction.

7 0

3 years ago

60=-5(-6-3z) <br>Find the solution

Sophie [7]

Answer:

z=2

Step-by-step explanation:

60=-5(-6-3z)

60= 30+15z

60-30=15z

30/15=15z/15

2=z

6 0

3 years ago