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tigry1 [53]
3 years ago
11

I need help with this slope problem​

Mathematics
1 answer:
sweet [91]3 years ago
8 0

Answer:

what is it? you did not post the picture

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Find the equation of the tangent line to the curve (a lemniscate)
olya-2409 [2.1K]

Answer:

m=\frac{9}{13} and b=\frac{40}{13}

Step-by-step explanation:

The equation of curve is

2(x^2+y^2)^2=25(x^2-y^2)

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})

4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})

The point of tangency is (-3,1). It means the slope of tangent is \frac{dy}{dx}_{(-3,1)}.

Substitute x=-3 and y=1 in the above equation.

4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})

40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})

-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}

80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240

130\frac{dy}{dx}=90

Divide both sides by 130.

\frac{dy}{dx}=\frac{9}{13}

If a line passes through a points (x_1,y_1) with slope m, then the point slope form of the line is

y-y_1=m(x-x_1)

The slope of tangent line is \frac{9}{13} and it passes through the point (-3,1). So, the equation of tangent is

y-1=\frac{9}{13}(x-(-3))

y-1=\frac{9}{13}(x)+\frac{27}{13}

Add 1 on both sides.

y=\frac{9}{13}(x)+\frac{27}{13}+1

y=\frac{9}{13}(x)+\frac{40}{13}

Therefore, m=\frac{9}{13} and b=\frac{40}{13}.

5 0
2 years ago
william earns$13 an hour working.last week he worked h hours and three times as many hours. write an expression for the amount o
drek231 [11]
13 x h = A or answer

4 0
2 years ago
 The distance between city A and B is 600 km. The first train left A and headed towards B at the speed of 60 km/hour. The secon
hoa [83]
\bf \begin{array}{ccccllll}
&distance&rate(km/hr)&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&60&t\\
B&600-d&v&t+3
\end{array}

\bf \textit{meaning}\implies 
\begin{cases}
d=(60)(t)
\\ \quad \\
600-d=(v)(t+3)\\
------------\\
d=\boxed{60t}\qquad thus
\\ \quad \\
600-\boxed{60t}=v(t+3)\leftarrow \textit{solve for "t"}
\end{cases}

keep in mind, that "t" is the time when the train at A station, left towards B station

they met, at some time "t", and by the time that happened, train from A
which started 3 hours earlier, had already covered "d" distance,
whatever that is
and the train coming from B, covered, 600-d, or the difference
8 0
2 years ago
Find the perimeter of a rectangle with a length of 6x+3 and a width of -2x-5.
blsea [12.9K]
The simple way of calculating a perimeter is to add all of the sides.
For a rectangle this would equal 2 x length + 2 x width
P = 2(6x + 3) + 2(-2x - 5)
= 12x + 6 - 4x - 10
= 8x - 4

At a higher level both the length and width must be greater than zero (= zero is a trivial rectangle)
6x + 3 > 0
6x > -3
x > -0.5

-2x - 5 > 0
2x + 5 < 0 (multiplying by -1 reverses the inequality)
2x < -5
x < -2.5

This rectangle cannot exist as x cannot be < -2.5 and > -0.5 at the same time!
8 0
3 years ago
Read 2 more answers
Estimate the sum 8/10 + 1/9
RideAnS [48]

Answer:

41/45

Step by Step Explanation:

Add: 8/

10

+ 1/

9

= 8 · 9/

10 · 9

+ 1 · 10/

9 · 10

= 72/

90

+ 10/

90

= 72 + 10/

90

= 82/

90

= 2 · 41/

2 · 45

= 41/

45

For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(10, 9) = 90. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 10 × 9 = 90. In the following intermediate step, cancel by a common factor of 2 gives 41/

45

.

In other words - eight tenths plus one ninth = forty-one forty-fifths.

8 0
2 years ago
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