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3 years ago

Write the point-slope for the seven following equations

1 answer:

7 0

K I will try my best

1) y+3=7/3(x+3)

2) y-0=3/4(x+4)

3) y-4=-5(x+1)

4) y-0=1/3(x-2)

5) y+4=-1(x-4)

6) y-4=-5/2(x+2)

7) y+1=-1/2(x+4)

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Can someone solve this for me

Alecsey [184]

Answer:

$12 {y}^{9} - 6 {y}^{5} + 4 {y}^{2} + 21$

Step-by-step explanation:

divide each term by 2y^3

Multiply through by the least common denominator.

3 0

3 years ago

Find the lateral area of a cylinder whose radius is 6 cm and height is 20 cm.

zaharov [31]

The surface area of a cylinder is define by the formula S.A.=2πrh+2πr^2, where the first part of the formula refers to the lateral area, perimeter, or circumference and the second part to the area of the bases, which are circles.

On this exercise it is asked to find the lateral area of a cylinder whose radius is 6 cm, and has a height of 20cm. To find the lateral area of the cylinder you should substitute this values into the formula, S.A.=2πrh, and as can be seen the answers are given in terms of π or pi.

S.A.=2πrh
S.A.=2π(6cm)(20cm)
S.A.=2π(120cm)
S.A.=240π cm^2

The lateral area of the cylinder is 240π cm^2 or in other words letter B from the given choices.

7 0

3 years ago

Read 2 more answers

Can someone help please

Phoenix [80]

9514 1404 393

Answer:

16x +8h

Step-by-step explanation:

Use the given function arguments and simplify.

$\dfrac{f(x+h)-f(x)}{h}=\dfrac{8(x+h)^2-8x^2}{h}=\dfrac{8(x^2+2hx+h^2)-8x^2}{h}\\\\=\dfrac{8x^2 +16xh+8h^2-8x^2}{h}=\dfrac{16xh+8h^2}{h}=\boxed{16x +8h}$

7 0

3 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

8 0

3 years ago

If i make 1.5 million dollars per day, plus an extra 250k interest every 31 hours, how much money do i make per day?

lidiya [134]

Answer: Alot

Step-by-step explanation:

4 0

3 years ago