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AVprozaik [17]
3 years ago
11

HELP ASAP !

Mathematics
1 answer:
AnnZ [28]3 years ago
4 0

Answer:

the for the question is

a= -5

b= -4

c= 4

d= -1

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The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at
inna [77]

Answer:

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Step-by-step explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

\dfrac{5 \cdot x - 2}{5} = \dfrac{5 \cdot x - 2}{x^2 + 1}

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

\dfrac{5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8}{5} = 0

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5  = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve, \dfrac{\mathrm{d} y}{\mathrm{d} x}, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

\dfrac{\mathrm{d} y}{\mathrm{d} x}= \dfrac{\mathrm{d} \left (\dfrac{5 \cdot x - 2}{x^2 + 1}  \right )}{\mathrm{d} x} = \dfrac{5\cdot \left ( x^{2} +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x}{\left (x^2 + 1 ^{2} \right )}.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.92

At x = 2/5, \dfrac{\mathrm{d} y}{\mathrm{d} x} = 4.3

At x = 2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.

7 0
3 years ago
The following group of values was entered into the TVM solver on a graphing calculator
Kisachek [45]

Based on the entries into the TVM solver, the expression that will return the same value for PV is \frac{(620) (( 1 + 0.066)^{108} - 1) }{(0.066) (1 + (0.066))^{108}}.

<h3>Which formula will give the same Present Value?</h3>

The formula for present value for this series of cashflows is:

= \frac{(Payment) (( 1 + Interest rate)^{Number of years} - 1) }{(Periodic interest) (1 + (Periodic interest))^{Number of years}}

Based on the P/Y and the C/Y being 12, the number of periods given in the graphing calculator will be the same in the formula, which is 108 periods.

The interest rate of 6.6% will also be the same in the formula.

The best option is therefore <u>option B </u>which has all these figures and so will give the same present value.

Find out more on present value at brainly.com/question/25792915.

#SPJ1

8 0
2 years ago
Read 2 more answers
5x-6y=22 in slope intercept
puteri [66]
Slope intercept form means that you have to solve for y.
Subtract 5x from both sides
5x - 6y - 5x = 22 - 5x
= -6y = 22 - 5x
Then divide by -6
-6y/-6 = 22/-6 - 5x/-6
= y = -22/6 + 5x/6
You can reduce the fraction -22/6 to -11/3
y = -11/3 + 5x/6
Then swap the order of your terms so it looks nicer and you don't have a negative in front
y = 5x/6 - 11/3
4 0
3 years ago
Find the sum<br> 3+5+7+9…., n=14
natka813 [3]
24! If that is not correct please let me know
4 0
2 years ago
Read 2 more answers
This is for AP CALCULUS. I really need help on this. I’m trying to find the first and second derivative of these three functions
ale4655 [162]

Review Material:

\sqrt[n]{x^m} = x^{\frac{m}{n}}

\frac{d}{dx}[x^n]=nx^{n-1}\\

\frac{d}{dx}[\sin(x)]=\cos(x)\\

 \frac{d}{dx}[\cos(x)]=-\sin(x)\\

\frac{d}{dx}[constant]=0\\

Step-by-step explanation:

(a)

F(x) = -2\cos(x) +x^{\frac{4}{3}} -3e\\ \text{Note: 3e is a constant}\\F'(x) = 2\sin(x) + \frac{4}{3}x^{\frac{1}{3}}\\F''(x) = 2\cos(x) + \frac{4}{9}x^{-\frac{2}{3}}

(b)

F(x) = x^{-3} + \frac{1}{2}x^2-\sin(x)\\F'(x)=-3x^{-4} + x - \cos(x)\\F''(x) = 12x^{-5} + 1 + \sin(x)

(c)

F(x) = 2x^{-3}+x^{\frac{3}{4}}-4x\\F'(x) = -6x^{-4} + \frac{3}{4}x^{-\frac{1}{4}}-4\\F''(x)= 24x^{-5} - \frac{3}{16}x^{-\frac{5}{4}}

8 0
2 years ago
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