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3 years ago

What is a zero of the problem

Mathematics

1 answer:

amm18123 years ago

4 0

The answer is C hope this would help you

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If xy + y² = 6, then the value of dy/dx at x= -1 is<br>

mylen [45]

Hi there!

$\large\boxed{\text{At (-1, -2), }\frac{dy}{dx} = -\frac{2}{5}}}$

$\large\boxed{\text{At (-1, 3), }\frac{dy}{dx} = -\frac{3}{5}}}$

We can calculate dy/dx using implicit differentiation:

xy + y² = 6

Differentiate both sides. Remember to use the Product Rule for the "xy" term:

(1)y + x(dy/dx) + 2y(dy/dx) = 0

Move y to the opposite side:

x(dy/dx) + 2y(dy/dx) = -y

Factor out dy/dx:

dy/dx(x + 2y) = -y

Divide both sides by x + 2y:

dy/dx = -y/x + 2y

We need both x and y to find dy/dx, so plug in the given value of x into the original equation:

-1(y) + y² = 6

-y + y² = 6

y² - y - 6 = 0

(y - 3)(y + 2) = 0

Thus, y = -2 and 3.

We can calculate dy/dx at each point:

At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.

At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.

5 0

3 years ago

Match the fractions that are equivalent to each other?

MrRissso [65]

Answer:

1&a 2&b 3&c 4&d

Step-by-step explanation:

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4 0

2 years ago

Read 2 more answers

Please help! A-level maths.

astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.