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Alja [10]
3 years ago
8

Are these the correct answers? I hunk I got some wrong pls help

Mathematics
1 answer:
klio [65]3 years ago
3 0

yes you got them right i believe. i worked them out just a second ago and i got the same answers. i am sorry if i got them wrong. if you need to show the work then you should do that.

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I need help with a math question, ASAP!! Please, I will give Brainliest. Below is a screenshot of the math problem
Dmitrij [34]
Y = 6x
4x + y = 7

(substitute y = 6x in the ‘y’ spot in 4x + y = 7)
4x + 6x = 7
10x = 7
/10 on both sides
x = 7/10

(substitute x = 7/10 in the ‘x’ spot in either of the two equations)
y = 6x
y = 6(7/10)
y = 4.2 = 21/5

Answer:
x = 7/10
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7 0
3 years ago
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Look at the figure: An image of a right triangle is shown with an angle labeled x. If tan x° = e divided by 8 and sin x° = e div
Nastasia [14]

Answer:

cos x = 8/f.

Step-by-step explanation:

We can use the identity :

tan x = sin x / cos x.

Substituting:

e/8 = e/f / cos x

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cos x = e/f * 8/e

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8 0
3 years ago
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The point C(5, 4) is translated 9 units down.
Naily [24]

Answer:

The new point is at (5, -5)

Step-by-step explanation:

When a point is being translated down, they are being subtracted on the y-value. So all you have to do is subtract 9 from 4 and we end up with -5, our new y-value.

(5, -5)

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3 years ago
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A projectile is fired with muzzle speed 220 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe
Allushta [10]

Answer:

  • 4968.6 m from where it was fired
  • 221.33 m/s

Step-by-step explanation:

For the purpose of this problem, we assume ballistic motion over a stationary flat Earth under the influence of gravity, with no air resistance.

We can divide the motion into two components, one vertical and one horizontal. For muzzle speed s and launch angle θ, the horizontal speed is presumed constant at s·cos(θ). The initial vertical speed is then s·sin(θ) and the (x, y) coordinates as a function of time are ...

  (x, y) = (s·cos(θ)·t, -4.9t² +s·sin(θ)·t + h₀) . . . . . where h₀ is the initial height

To find the range, we can solve the equation y=0 for t, and use this value of t to find x.

Using the quadratic formula, we find t at the time of landing to be ...

  t = (-s·sin(θ) - √((s·sin(θ))²-4(-4.9)(h₀)))/(2(-4.9))

  t = (s/9.8)(sin(θ) +√(sin(θ)² +19.6h₀/s²))

For s = 220, θ = 45°, and h₀ = 30, the time of flight is ...

  t ≈ 31.939 seconds

Then the horizontal travel is

  x = 220·cos(45°)·31.939 ≈ 4968.6 . . . . meters

__

As it happens, the value under the radical in the above expression for time, when multiplied by s, is the vertical speed at landing. The horizontal speed remains s·cos(θ), so the resultant speed is the Pythagorean sum of these:

  landing speed = s·√(cos(θ)² +sin(θ)² +19.6h₀/s²) ≈ s√(1 +0.012149)

  ≈ 221.33 m/s

_____

Note that the landing speed represents the speed the projectile has as a consequence of the potential energy of its initial height being converted to kinetic energy that adds to the kinetic energy due to its initial muzzle velocity.

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3 years ago
Porfa ayuda plissss<br>(SINO SABES NO CONTESTES)<br>DOY PUNTOS​
Leto [7]

Answer:

B

Step-by-step explanation:

Si cuentas desde el once hasta el espacio entre A y B llegas hasta 5. Entonces el número que está en ese espacio es 11.05

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3 years ago
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