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3 years ago

Help me this is hard

Mathematics

1 answer:

siniylev [52]3 years ago

3 0

Following PEMDAS is crucial to solving these. Please Excuse My Dear Aunt Sally

P - Parenthesis

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

Multiplication/Division and Addition/Subtraction are interchangeable.

Now then, number 1 has the following:

$4-2*409-8*96$

There are no parenthesis or exponents, but there is multiplication, so we will start with multiplying. There are two multiplication expressions in the problem.

$-2 * 409 = -818$

$-8 * 96 = -768$

Since you did that, your answer has been simplified to:

$4-818-768$

Now, all you have to do is combine your like terms. Since every term is alike, you can combine the whole expression.

$4-818-768 = -1582$

So, your final answer would be:

$4-2*409-8*96 = -1582$

Hopefully with this information, you can solve the rest. If you have any questions, let me know.

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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.

6 0

2 years ago

Which angles are complementary to each other, will give brainliest

gayaneshka [121]

3 nd 4

Step-by-step explanation:

hope this helps

8 0

2 years ago

1/3 (12 - 24v) = -2(4v -2)

ANEK [815]

Opening brackets gives;
4-8v=-8v+4
Putting like terms together;
4-4=-8v+8v
0=0
There is no solution thus.