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2 years ago

Which equation is represented by the table

Mathematics

1 answer:

lyudmila [28]2 years ago

4 0

Answer:

B. b = 3a + 2

Step-by-step explanation:

We can write the equation in slope-intercept form as b = ma + c, where,

m = slope/rate of change

c = y-intercept/initial value

✔️Find m using any two given pair of values, say (2, 8) and (4, 14):

Rate of change (m) = change in b/change in a

m = (14 - 8)/(4 - 2)

m = 6/2

m = 3

✔️Find c by substituting (a, b) = (2, 8) and m = 3 into b = ma + c. Thus:

8 = 3(2) + c

8 = 6 + c

8 - 6 = c

2 = c

c = 2

✔️Write the equation by substituting m = 3 and c = 2 into b = ma + c. Thus:

b = 3a + 2

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2 years ago

If the radius of the circle is 3 in, what is the area of the circle?

murzikaleks [220]

The area of the circle is 28.26 in²

Explanation:

Given:

Radius of the circle, r = 3 in

Area of circle, A = ?

We know:

Area of circle = πr²

The value of π is 3.14

Thus,

A = 3.14 X (3)²

A = 3.14 X 9

A = 28.26 in²

Therefore, the area of the circle is 28.26 in²

3 0

2 years ago

I need help with this proof!

mr Goodwill [35]

Answer:

∠B ≅ ∠F ⇒ proved down

Step-by-step explanation:

In the two right triangles, if the hypotenuse and leg of the 1st right Δ ≅ the hypotenuse and leg of the 2nd right Δ, then the two triangles are congruent

Let us use this fact to solve the question

→ In Δs BCD and FED

∵ ∠C and ∠E are right angles

∴ Δs BCD and FED are right triangles ⇒ (1)

∵ D is the mid-point of CE

→ That means point D divides CE into 2 equal parts CD and ED

∴ CD = ED ⇒ (2) legs

∵ BD and DF are the opposite sides to the right angles

∴ BD and DF are the hypotenuses of the triangles

∵ BD ≅ FD ⇒ (3) hypotenuses

→ From (1), (2), (3), and the fact above

∴ Δ BCD ≅ ΔFED ⇒ by HL postulate of congruency

→ As a result of congruency

∴ BC ≅ FE

∴ ∠BDC ≅ ∠FDE

∴ ∠B ≅ ∠F ⇒ proved

3 0

2 years ago

(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

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

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

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

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

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

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

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

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

if the rate of change for one linear function is positive and for another is negative can they both be either increasing or decr

Alla [95]

Answer:

They can never be both either increasing or decreasing.

Step-by-step explanation:

If the rate of change i.e. the slope of one linear function is positive, that means the graph of the linear function makes angle which varies between 0° to 90° with respect to the positive direction of the x-axis.

Therefore, the function must be increasing.

Again, if the rate of change i.e. the slope of one linear function is negative, that means the graph of the linear function makes angle which varies between 90° to 180° with respect to the positive direction of the x-axis.

Therefore, the function must be decreasing.

Hence, if the rate of change of one linear function is positive and for another is negative, they can never be both either increasing or decreasing. (Answer)

6 0

2 years ago