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drek231 [11]
3 years ago
8

Harry is trying to solve the equation y = 2x^2 − x − 6 using the quadratic formula. He has made an error in one of the steps bel

ow. Find the step where Harry went wrong.

Mathematics
1 answer:
Bond [772]3 years ago
4 0
He made a mistake in step #2. It seemed to be a trivial mistake because it involved signs, but it still had a great impact. Since step#2, his solution was already wrong.

Instead of
(-1)²-4(2)(-6) = 1 + 48 = 49

What he did is
(-1)²-4(2)(-6) = -1 + 48 = 47
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A cylinder and a cone have the same diameter: 8 inches. The height of the cylinder is 9 inches. The height of the cone is 18 inc
Elan Coil [88]
\sf Cylinder ~volume = \pi r^2 h \\  \\ v = \pi (4)^2(9) \\  \\ v=\pi (16)(9) \\  \\ v=144 \pi  \\ v=452.16\\  \\  \\ Cone~volume = \frac{1}{3} \pi r^2 h \\  \\ v =  \frac{1}{3} \pi (4)^2 (18) \\  \\ v=96 \pi \\ v=301.44 \\  \\  \\ The~ volume ~of ~a ~cylinder ~is~ 1.5~ times~ more ~than ~the ~volume ~of~a~cone.
4 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
I really need help what's the answer??
lys-0071 [83]
I think you just have to add all of the totals up??
5 0
3 years ago
Name an inscribed angle
Vladimir79 [104]

Answer:

BHF

Step-by-step explanation:

Definition of inscribed

6 0
3 years ago
1. The diameter of a sphere is 21.6 cm. The volume of the sphere is __
sveta [45]
Detailed Answers:

Volume of a Sphere (V) = 4/3 πr^3

1. Diameter (d) = 21.6 cm
Radius (r) = 21.6/2 = 10.8 cm

Therefore,
= 4/3 πr^3
= 4/3 * 22/7 * (10.8)^3
= 4/3 * 22/7 * 1259.712
= 88/21 * 1259.712
=> 5278.79

Volume (V) = 5278.79 cm^3

2. Diameter (d) = 16 cm
Radius (r) = 16/2 = 8 cm

Therefore,
= 4/3 πr^3
= 4/3 * 22/7 * (8)^3
= 4/3 * 22/7 * 512
= 88/21 * 512
=> 2145.52

Volume (V) = 2145.52 cm^3

3. Diameter (d) = 24 cm
Radius (r) = 24/2 = 12 cm

Therefore,
= 4/3 πr^3
= 4/3 * 22/7 * (12)^3
= 4/3 * 22/7 * 1728
= 88/21 * 1728
=> 7241.14

Volume (V) = 7241.14 cm^3

4. Diameter (d) = 6 cm
Radius (r) = 6/2 = 3 cm

Therefore,
= 4/3 πr^3
= 4/3 * 22/7 * (3)^3
= 4/3 * 22/7 * 27
= 88/21 * 27
=> 113.14

Volume (V) = 113.14 cm^3
6 0
1 year ago
Read 2 more answers
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