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2 years ago

Systems of Equations:

1 answer:

4 0

$v_b-the\ speed\ of\ the\ boat\\v_c-the\ speed\ of\ the\ current\\\\v_b+v_c=\frac{210}{10}=21\ (mph)\\\\v_b-v_c=\frac{210}{70}=3\ (mph)\\\\ +\left\{\begin{array}{ccc}v_b+v_c=21\\v_b-v_c=3\end{array}\right\\-----------\\.\ \ \ \ \ \ 2v_b=24\ \ \ \ /:2\\.\ \ \ \ \ \ \ \ \ v_b=12\ (mph)\\\\12+v_c=21\\v_c=21-12\\v_c=9\ (mph)$

$Answer:\\the\ speed\ of\ the\ both\ is\ 12mph\\the\ speed\ of\ the\ current\ is\ 9mph$

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What is the slope of the line that passes through the points (-9, -3) and (1, -3)?

anastassius [24]

Answer:

The slope is 0.

Step-by-step explanation:

m= (y2-y1)/(x2-x1).

y2 is -3, y1 is -3.

x2 is 1 and x1 is -9.

m= -3 -(-3)/ 1- (-9)

m= 0/10

m=0

8 0

2 years ago

Which strategies can be used to solve this problem?

vladimir2022 [97]

You want total number made in 4 m onths, so adding all numbers is correct, but it is per 1 month, so multiply by 2

A because you need brain for most problems
D works

pick A and D

8 0

2 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

2 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the dista

Kamila [148]

40mph for Car B I think. 45mph for Car A. 45(mph) times 4(h) is 180(distance) and 40 times 4.5 is also 180. Sorry if I'm wrong.

8 0

2 years ago

50 POINTS!!!

san4es73 [151]

Step-by-step explanation:

<u>Step 1: Determine an ordered pair</u>

A solution of an equation just means that the point lies on the line. We can find any y-value when we plug in a specific x-value. For example, if we want to know what ordered pair lies at x=1, we just plug in y = -1/2(1) and solve for y which gives us -1/2. This gives us an ordered pair of (1, -1/2). We can continue to do this for any x value.

We can also reverse the order and plug in the y-value and get the x-value in order to accomplish the same goal but it's a bit harder.

Hope this helps!

6 0

1 year ago

Read 2 more answers