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Goryan [66]
3 years ago
12

HELP: due sep 18!!

Mathematics
1 answer:
damaskus [11]3 years ago
4 0

Answer:

The correct order that the machines should be stacked so that when 15 dropped into the first machine, and all four machines have had their effect, the last machine's output is -6 can be written as follows;

4th machine → 2nd machine → 1st machine → 3rd machine

Step-by-step explanation:

Given that the initial input is 15 and the final output is -6, we have;

For the fourth machine;

y = -2·x + 34 = -2×15 + 34 = 4

The output is 4

Next to the second machine

y = (x - 2)²

The input is x = 4, therefore;

y = (4 - 2)² = 2² = 4

The output is 4

Next to the first machine

y  = - \left | 3\cdot x \right |

The input, x = 4, therefore;

y  = - \left | 3\times 4 \right | = -12

The output is -12

Next to the third machine

y = -x/3 -10

The input is x = -12, therefore;

y = -(-12)/3 -10 = 12/3 -10 = 4 - 10 = -6

The output is -6.

Therefore, the correct order that the machines should be stacked so that when 15 dropped into the first machine, and all four machines have had their effect, the last machine's output is -6 can be written as follows;

4th → 2nd → 1st → 3rd

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20 kg remaining in bag 1.
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3 years ago
H(t) = -10+ - 6<br>h()= -44​
lesya [120]

Answer:

-44

Step-by-step explanation:

4 0
3 years ago
MATH HELP!!! 100PTS!!!
Lostsunrise [7]

Answer:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Step-by-step explanation:

<u>Given functions</u>:

  p(x)=2x-4

  r(x)=\dfrac{6x-1}{9x+1}

Solve for p(x) = r(x):

\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=18, \quad b=-40, \quad c=-3

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}

Therefore, the solutions are:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

6 0
2 years ago
Which statements are always true regarding the diagram? Check all that apply. m∠3 + m∠4 = 180° m∠2 + m∠4 + m∠6 = 180° m∠2 + m∠4
Nostrana [21]

Answer:

The true statements are:

m∠ 3 + m∠ 4 = 180° ⇒ 1st

m∠ 2 + m∠ 4 + m∠ 6 = 180° ⇒ 2nd

m∠ 2 + m∠ 4 = m∠ 5 ⇒ 3rd

Step-by-step explanation:

* Look to the attached diagram to answer the question

# m∠ 3 + m∠ 4 = 180°

∵ ∠ 3 and ∠ 4 formed a straight angle

∵ The measure of the straight angle is 180°

∴ m∠ 3 + m∠ 4 = 180° ⇒ <em>true</em>

# m∠ 2 + m∠ 4 + m∠ 6 = 180°

∵ ∠ 2 , ∠ 4 , ∠ 6 are the interior angles of the triangle

∵ The sum of the measures of interior angles of any Δ is 180°

∴ m∠ 2 + m∠ 4 + m∠ 6 = 180° ⇒ <em>true</em>

# m∠ 2 + m∠ 4 = m∠ 5

∵ In any Δ, the measure of the exterior angle at one vertex of the

  triangle equals the sum of the measures of the opposite interior

  angles of this vertex

∵ ∠ 5 is the exterior angle of the vertex of ∠ 6

∵ ∠2 and ∠ 4 are the opposite interior angles to ∠ 6

∴ m∠ 2 + m∠ 4 = m∠ 5 ⇒ <em>true </em>

# m∠1 + m∠2 = 90°

∵ ∠ 1 and ∠ 2 formed a straight angle

∵ The measure of the straight angle is 180°

∴ m∠1 + m∠2 = 90° ⇒ <em>Not true</em>

# m∠4 + m∠6 = m∠2

∵ ∠ 4 , ∠ 6 , ∠ 2 are the interior angles of a triangle

∵ There is no given about their measures

∴ We can not says that the sum of the measures of ∠ 4 and ∠ 6 is

  equal to the measure of ∠ 2

∴ m∠4 + m∠6 = m∠2 ⇒ <em>Not true</em>

<em></em>

# m∠2 + m∠6 = m∠5

∵ ∠ 5 is the exterior angle at the vertex of ∠ 6

∴ m∠ 2 + m∠ 6 = m∠ 5 ⇒ <em>Not true</em>

6 0
3 years ago
Read 2 more answers
One of the roots of the quadratic equation 4mnx^2 – 6m^2x – 6n^2x + 9mn = 0 (m, n ≠ 0) is a)-3m/2n b)3m/2n c)2m/3n d)-2m/3n
maxonik [38]

Answer:

B.\ \frac{3m}{2n}

Step-by-step explanation:

Given

4mnx^2 - 6m^2x - 6n^2x + 9mn = 0\ (m, n \neq  0)

Required

Calculate one of the root of the equation

4mnx^2 - 6m^2x - 6n^2x + 9mn = 0

Factorize

2mx(2nx - 3m) -3n(2nx - 3m) = 0

(2mx - 3n)(2nx - 3m) = 0

Split equation

2mx - 3n = 0\ or\ 2nx - 3m = 0

Make x the subject of formula in both expressions

2mx = 3n\ or\ 2nx = 3m

x = \frac{3n}{2m}\ or\ x = \frac{3m}{2n}

<em>From the list of given options, one of the roots of the equation is</em> \frac{3m}{2n}

4 0
3 years ago
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