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Masja [62]
3 years ago
15

Acute, obtuse, and right are ways in which you can classify triangles by their ___.

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
7 0

Answer

A.

Step-by-step explanation:

Acute obtuse and right are all types of angles

mote1985 [20]3 years ago
4 0

Answer:

A) angles

Step-by-step explanation:

it is this because they r called right angels. not right sides

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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
PLS ANSWER BEST ANSWER GETS BRAINLIEST Tim wants to decorate a rectangular card with a width of 1/2 inch and an area of 7/4 squa
vovikov84 [41]

Answer:

\frac{7}{2}in

Step-by-step explanation:

First you must set up an equation. \frac{1}{2}x=\frac{7}{4}

Then, you must solve for x.          x=\frac{7}{2}

6 0
3 years ago
Camille said 4/5is equivalent to 24/30. Check her work by making a table of equivalent ratios
Nataly [62]

<em>The five equivalent fractions of 4/5 are: </em>

<em> 4 × 2/5 × 2 = 8/10 </em>

<em> 4 × 3/5 × 3 = 12/15 </em>

<em> 4 × 4/5 × 4 = 16/20 </em>

<em> 4 × 5/5 × 5 = 20/25 </em>

<em> 4 × 6/5 × 6 = 24/30</em>

<em />

<em>8 : 10</em>

<em>12 : 15</em>

<em>16 : 20</em>

<em>20 : 25</em>

<em>24 : 30</em>

3 0
3 years ago
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