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3 years ago

At which points are the equations y = x2 + 5x − 2 and y = x + 1 equal?

Mathematics

1 answer:

klemol [59]3 years ago

5 0

Answer:

When they are equal, y=y, so we can say:

x^2+5x-2=x+1 subtract x from both side

x^2+4x-2=1 add 2 to both sides

x^2+4x=3, now add half the linear coefficient squared, (4/2)^2=4

x^2+4x+4=3+4 now the left side is a perfect square

(x+2)^2=7 now take the square root of both sides

x+2=±√7 subtract 2 from both sides

x=-2±√7

x=-2+√7 and -2-√7

y=x+1 the two points where these equations are equal are:

(-2-√7, -1-√7) and (-2+√7, -1+√7)

or approximately:

(-4.65, -3.65) and (0.65, 1.65)

Step-by-step explanation:

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natima [27]

Answer:

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3 years ago

Help (this for a friend)<br> a 62.1<br> b 117.4<br> c 122.1<br> d 164.5

kramer

Answer:

c. 122.1 cm

Step-by-step explanation:

so l * w + $\frac{\pi r^2}{2}$

since 6 cm is the diameter

6/2 = 3

the radius (or r) is 3

so 18 * 6 + $\frac{3.14 * 3*3}{y}$

108 + $\frac{3.14*9}{y}$

108 + $\frac{28.26}{2}$

108 + 14.13 = 122.13

simplified: 122.1 cm

hope this helps:)

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2 years ago

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Whay 3 tims 99 Pls help confused thanks

timurjin [86]

3 times 99 is 297...

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3 years ago

Read 2 more answers

Select the values that make the inequality -c>-4 true.

Vera_Pavlovna [14]

Answer:

c < 4

Step-by-step explanation:

whenever you divide or multiply an inequality by a negative it is a rule that you must reverse the inequality symbol

so -c > -4 means you must divide or multiply by a negative one to get 'c' by itself; therefore it becomes c < 4

6 0

3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago