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2 years ago

11

Bridge math sem 1 .! Need helpp.!!

1 answer:

Ray Of Light [21]2 years ago

8 0

The answer should be C

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Write as a numerical expression: “ Six less than twice a number is four “

VladimirAG [237]

Answer:

2x - 6 = 4

Step-by-step explanation:

Firstly, let's assume the number is x

Then, twice the number is 2 times x = 2x

6 less than twice the number is four

2x - 6 = 4

2x - 6 + 6 = 4 + 6

2x = 10

2x/2 = 10/2

x = 5

6 0

2 years ago

What is the domain of this exponential function? y=2x

trasher [3.6K]

Answer:

x∈R

Step-by-step explanation:

$y=2^{x}$

2>0

The domain is all real numbers.

x∈R

8 0

3 years ago

Can someone help me with the middle question?

kotegsom [21]

You should do 1.0 divided by 7 and the answer would be 7

7 0

3 years ago

A total of 60 students in your cousin's school are taking Spanish class this year. This is 25% more than the number of students

alina1380 [7]

Since 25% is 1/4, you multiply 60 by 1/4 to get how many students took Spanish class last year. So, 60 times 1/4 is 60/4, which is 15. Then you subtract 15 from 60 to get 45. So, 45 students took Spanish class last year.

7 0

2 years ago

Read 2 more answers

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

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

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

5 0

2 years ago