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3 years ago

Which table represents a function?

Mathematics

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

the first table represents a function

Step-by-step explanation:

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The measure of ∠ABD is (0.12x+68)° and the measure of ∠CBD is (0.13x+24)°. Find the value of x.

kondor19780726 [428]

Answer:

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Step-by-step explanation:

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3 years ago

Find the amount of money accumulated after investing a principle P for years t at interest rate r, compounded continuously. P =

Nitella [24]

Answer:

The accumulated amount would be $4,353.56

Step-by-step explanation:

r= 0.08

A= 3, 200(1+0.08)^4

= 3,200(1.08)^4

= $4,353.56

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3 years ago

Look at the attachment! This is algebra. 10 points!

boyakko [2]

If x∆y = 3x - y², then

5∆1 = 3×5 - 1² = 15 - 1 = 14

and

14∆6 = 3×14 - 6² = 42 - 36 = 6

So (5∆1)∆6 = 6.

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2 years ago

Help with this math question, I am graphing them to see which one is correct, but please help so I can speedy out of this, pleas

sergejj [24]

Answer:

Last one: 3x²

Step-by-step explanation:

Quadratics are functions with the power of 2 as the highest.

Option 1 is a Reciprocal function

Option 2 is a linear function

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6 0

2 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

3 years ago