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3 years ago

9-3÷ 1/3 + 1= plz answer correctly because i can get grounded and I will give brainliest and give 20 points

Mathematics

2 answers:

fgiga [73]3 years ago

6 0

The answer should be 1

Bumek [7]3 years ago

3 0

The answer is 19.0000002
explanation: 9-3 is 6
turn 1/3 into a decimal
6 divided by 1/3 is 18.0000002
18.0000002 + 1 is 19.0000002

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(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 :

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

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

10($6.75)+7($1.99)+$15.00=$196.43

34kurt

96/43 = 196/43
the equation is false.
Hope this helps!

5 0

3 years ago

Emily is saving $300 a month for a $9,000 down payment on a new car. she will have her down payment total in 18 months. she got

Burka [1]

It would fewer her months by 10

4 0

3 years ago

Read 2 more answers

One and one tenth decimal number

Elden [556K]

Hi there!

One and one tenth looks hard but that would be because it's in word form.
In standard, or decimal form, this would be 1.1.
The first 1 represents the whole number and the second 1 represents the tenths.

Hope this helps! Let me know if you need anymore help! :)

3 0

3 years ago

Read 2 more answers

Question 8 of 10 What is the median of this data set? 14, 18, 31, 34, 44, 50 A. 32.5 B. 33 C. 31.5 O O D. 31 SUBM

kozerog [31]

Answer:

32.5

Step-by-step explanation:

median is the middle number, in this case the sum of the two middle number divided by 2

31 + 34 = 65/2

32.5

8 0

3 years ago

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