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3 years ago

Find the missing length of the triangle. 26 in. a 24 in.

Mathematics

2 answers:

sleet_krkn [62]3 years ago

7 0

<h3>Answer: a = 10</h3>

Work Shown:

Apply the pythagorean theorem to get...

a^2 + b^2 = c^2

a^2 + 24^2 = 26^2

a^2 + 576 = 676

a^2 = 676 - 576

a^2 = 100

a = sqrt(100)

a = 10

Elis [28]3 years ago

3 0

Answer:

Step-by-step explanation:

26^2-24^2=100

sqrt of 100 is 10

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3. The product of 7/10 and another factor

kicyunya [14]

The correct answer is d 7/7

4 0

3 years ago

Write an equivalent expression to present patricia total earnings this week 7h + 135 + 5h = ?h + ?

Paul [167]

Answer: 12 h + 135

Step-by-step explanation: 7h +135 + 5h = ?h + ?

7h +5h +135

=12h +135

So u combine like terms together which are 7h and 5h which gives you 12 h.And then you have 135 left,so you add 12h to 135.

Ps.If you don't know what like terms are.I will explain down below.

Like terms are variable that are the same.For an example 4y and 3y are like terms .Another example is 1/3 xy and 6xy .

Hope this helps :) and if it does please give brainliest .

8 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$