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natta225 [31]
2 years ago
8

Solve the system by substitution. 4y = 5x -10y =-50

Mathematics
1 answer:
Trava [24]2 years ago
7 0

Answer:

x = -20 ; y =-5

Step-by-step explanation:

Eqn. 1 ----> 4y = x

Eqn. 2 ----> 5x-10y = -50

(Simplifying eqn.2 further)

=>5(x-2y)=-50

=>x-2y=\frac{-50}{5} =-10

(Substituting the value of x from eqn. 1)

=>4y-2y=-10

=>2y=-10

=>y =\frac{-10}{2} =-5

Now, substituting the value of y in eqn. 1 ,

x = 4*(-5)=-20

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X intercept = -1,  y = 0           (-1, 0)

<span>y intercept = 2,  x = 0           (0, 2)
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slope between (-1, 0) and (0, 2):

slope = (2 -0)/(0 - -1) =  2/1 = 2,           m = 2

Using point   (-1, 0)   x₁ = -1, y₁ = 0

y - y₁ = m(x - x₁)

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6 0
3 years ago
Okay i need some help fast
Troyanec [42]

Answer:

<em>Dimensions : 1 by 1 by 2 feet,</em>

<em>Volume ⇒ 2 ft^3</em>

Step-by-step explanation:

Here, I can only assume that this rectangular prism is assigned to have dimensions 1 foot by 1 foot by 2 feet;

I can tell that these types of boxes have height and width as the same lengths, so given that;

<em>Dimensions : 1 by 1 by 2 feet,</em>

Volume ⇒ 1 * 1 * 2 ( by definition ) ⇒ 2 ft^3

In other words, <em>Volume ⇒ 2 ft^3</em>

6 0
3 years ago
Simplify each expression
Veseljchak [2.6K]
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5 0
3 years ago
A recent study found that the average length of caterpillars was 2.8 centimeters with a
pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 2.8, \sigma = 0.7.

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{4 - 2.8}{0.7}

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

4 0
2 years ago
How much square units of 50 units
Alborosie

Answer:

7.07106781187

Step-by-step explanation:

if its just asking \sqrt50

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5 0
3 years ago
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