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3 years ago

(n² – 3n – 21) = (n - 7)

Mathematics

1 answer:

kakasveta [241]3 years ago

5 0

Answer: n = 2 ± 3 √ 2

Step-by-step explanation: Solve the equation for

by finding

, and

of the quadratic then applying the quadratic formula.

Exact Form:

√

Decimal Form:

6.24264068

…

−

2.24264068

…

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Answer:

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Hope this helps! :)

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3 years ago

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Sloan [31]

Answer:

k=5b/4−3/4

Step-by-step explanation:

3-3k+7k=5b

Isolate the variable by dividing each side by factors that don't contain the variable.

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2 years ago

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3 years ago

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

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3 years ago

Segments AB, EF, and CD intersect at point C, and angle ACD is a right angle. Find the value of g. Please help me

iogann1982 [59]

Answer:

37 degree

Step-by-step explanation:

AB = 180 it is a straight line

ACD = 90 degree

DE+ DCB are complimentary angles they add up to 90 they dont need to be next to each other.

DE = 53 deg

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3 years ago

(n² – 3n – 21) = (n - 7)​

(n² – 3n – 21) = (n - 7)