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3 years ago

These are questions i notice a pattern with, but don't know how to approach. I need help!.

Mathematics

1 answer:

Otrada [13]3 years ago

7 0

Answer:

The equestrian overbid the horse by $485.75.

It takes the workman 12 days.

Step-by-step explanation:

Each nail costs 2 times more than the previous nail. So this is a geometric sequence where the first term is 1 and the common ratio is 2.

a₁ = 1

r = 2

There are 5 nails per horseshoe, and 4 horseshoes on a horse, for a total of 20 nails. The sum of the first n terms of a geometric series is:

S = a₁ (1 − rⁿ) / (1 − r)

Plugging in values:

S = 1 (1 − 2²⁰) / (1 − 2)

S = 1,048,575

The nails cost 1,048,575 cents, or $10,485.75. This is more than the cost of the horse, so the equestrian did not make a good deal.

Like before, this is a geometric sequence where the first term is 1 and the common ratio is 2.

The sum of the first n terms is:

S = a₁ (1 − rⁿ) / (1 − r)

Plugging in values:

4095 = 1 (1 − 2ⁿ) / (1 − 2)

4095 = 2ⁿ − 1

4096 = 2ⁿ

n = 12

It takes the worker 12 days.

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3 years ago

The vertex of a parabola is at (8,-1)and its y-interpret is-17. Which function represents the parabola

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (8, - 1), thus

y = a(x - 8)² - 1

To find a substitute (0, - 17), the coordinates of the y- intercept into the equation.

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4 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

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The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

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3 years ago

Assuming the population variances are unknown and equal, you wish to test whether the Group 1 population mean is equal to the Gr

Fantom [35]

Answer:

$n_1+n_2-2$

Step-by-step explanation:

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