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3 years ago

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HELPPP PLEASE!!!!

1 answer:

ad-work [718]3 years ago

5 0

Answer: B.f(n)9n-12

Step-by-step explanation:

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Please help me with these calculus bc questions

zhannawk [14.2K]

4. Compute the derivative.

$y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1$

Find when the gradient is 7.

$4x - 1 = 7 \implies 4x = 8 \implies x = 2$

Evaluate $y$ at this point.

$y = 2\cdot2^2-2-1 = 5$

The point we want is then (2, 5).

5. The curve crosses the $x$ -axis when $y=0$ . We have

$y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4$

Compute the derivative.

$y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}$

At the point we want, the gradient is

$\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}$

6. The curve crosses the $y$ -axis when $x=0$ . Compute the derivative.

$\dfrac{dy}{dx} = 3x^2 - 4x + 5$

When $x=0$ , the gradient is

$\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}$

7. Set $y=5$ and solve for $x$ . The curve and line meet when

$5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92$

Compute the derivative (for the curve) and evaluate it at these $x$ values.

$\dfrac{dy}{dx} = 4x + 7$

$\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}$

$\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}$

8. Compute the derivative.

$y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b$

The gradient is 8 when $x=2$ , so

$2a\cdot2 + b = 8 \implies 4a + b = 8$

and the gradient is -10 when $x=-1$ , so

$2a\cdot(-1) + b = -10 \implies -2a + b = -10$

Solve for $a$ and $b$ . Eliminating $b$ , we have

$(4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}$

so that

$4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}$ .

5 0

2 years ago

If anyone can plz help

8090 [49]

the picture is NOT loading.

4 0

3 years ago

Help me please. Your the best if you help.

ASHA 777 [7]

Answer: x = 25, y=15, z = 58

Step-by-step explanation:

4 0

1 year ago

Maria drove from Los Angeles (elevation 330 feet) to Death Valley (elevation –282 feet). What is the difference in elevation bet

motikmotik

Answer:

612 feet

Step-by-step explanation:

LA is located at 330 feet ABOVE SEA LEVEL

Death Valley is located 282 feet BELOW SEA LEVEL

We let the sea level be at 0 (consider a number line).

So,

LA would be at +330 feet

and

Death Valley would be at -282 feet

The elevation change between the two would be the difference:

330 - (-282) = 330 + 282 = 612 feet

The difference in elevation = 612 feet

3 0

3 years ago

Kayleigh decided to donate 30% of her savings. If she has $238 in her savings account, about ow much will she donate

Brrunno [24]

She will donate 0.30*238, or approximately $71.40

5 0

3 years ago