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2 years ago

PLS HELP ASAP ITS FOR MY ASSIGNMENT DUE RN!!! WHOEVER ANSWERS FIRST GETS BRAINLIEST, 5 STARS, AND A THANKS!!!!

Mathematics

2 answers:

Yuri [45]2 years ago

6 0

Nico added 1 1/2 instead of subtracting
1 1/2 should be subtracted from 9 7/11.

$\\ \sf\longmapsto 9\dfrac{7}{11}-(5+1\dfrac{1}{2})$

$\\ \sf\longmapsto4 \dfrac{7}{11}-1\dfrac{1}{2}$

$\\ \sf\longmapsto 4\dfrac{14}{22}-1\dfrac{11}{22}$

$\\ \sf\longmapsto 3\dfrac{3}{22}$

bonufazy [111]2 years ago

4 0

Answer:

A. He should subtract 1 1\2 instead of adding it

B. he should subtract 1 1\2 instead

C. the answer should be 3 3\22

Step-by-step explanation:

4 7\11 - 1 1\2

4 14\22 - 1 11\22

= 3 3\22

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The graph of the equation y=7x-3 has what slope?

Nikitich [7]

The slope of the line would be 7

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4 years ago

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Please help me with this math question will mark brainliest

LenaWriter [7]

Answer:

-9

Step-by-step explanation:

Using the sum/difference property of logarithms, we can rewrite the expression given as:

log b^3 + log c^3 - log √(a^3) --> log √(a^3) can also be written as log a^1.5

Next, we can use the power property of logarithms, and rewrite it again as:

3log b + 3log c - 1.5log a

Now, we can substitute the values of log a, log b, and log c:

3(11) + 3(-9) - 1.5(10)

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3 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

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

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

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3 years ago

Please help me i will give brainlist please...

Luda [366]

Answer: All i know is you have to multiply all of those numbers because its area

Step-by-step explanation:

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3 years ago

Will mark brainliest, pls help

elena-s [515]

Answer:

None of the answer options are correct

Step-by-step explanation:

cs + rh = w

cs = w - rh

s = (w - rh)/c

The second one could be corrected by exchanging c with 1/c

The last one would be correct with judiciously applied parentheses.

The first and third have too many errors to even bother trying to correct.

answer could also be written

s = w/c - rh/c

but it is not considered reduced form.

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2 years ago

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