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kow [346]
3 years ago
6

Solving systems by elimination 3x+y=9 y=3x+6

Mathematics
1 answer:
Svetach [21]3 years ago
5 0

\left\{\begin{array}{ccc}3x+y=9\\y=3x+6&\text{subtract 3x from both sides}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}3x+y=9\\-3x+y=6\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad2y=15\qquad\text{divide both sides by 2}\\.\qquad\qquad\boxed{y=7.5}\\\\\text{Put the value of y to the first equation}\\\\3x+7.5=9\qquad\text{subtract 7.5 from both sides}\\3x=1.5\qquad\text{divide both sides by 3}\\\boxed{x=0.5}\\\\Answer:\ \boxed{x=0.5\ and\ y=7.5\to(0.5,\ 7.5)}

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Step-by-step explanation:

4^m * 4^2 = 12

4^(m + 2) = 4^(log4 12)

m + 2 = log4 (12)

m = log4 (12) - 2.

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Before the eruption of a​ volcano, the volcano was 2.75 kilometers high. After the​ eruption, the volcano was 2.25 kilometers hi
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Before the eruption the vulcano was 2.75 km high, what is equal to (2,75x1000) 2750 m. After the eruption it was 2250 meters high. The difference would then be 2750-2250= 500 meters.
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4 years ago
The data below represent the weight, in kilograms, of 15 containers shipped from
Nikolay [14]

Answer:

(A) Maximum = 100 kg.

(B) Lower quartile = 55 kg.

(C) Minimum = 45 kg.

(D) Upper quartile = 93 kg.

(E) Median = 81 kg.

Step-by-step explanation:

A box-plot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

  1. Minimum (shown at the bottom of the chart)
  2. First Quartile (shown by the bottom line of the box)
  3. Median (or the second quartile) (shown as a line in the center of the box)
  4. Third Quartile (shown by the top line of the box)
  5. Maximum (shown at the top of the chart).

The data provided for the weight, in kilograms, of 15 containers shipped from  a factory in one week is as follows:

S = {45, 51, 53, 55, 55, 65, 75, 81, 84, 87, 93, 93, 95, 96, 100}

The data is already arranged in ascending order.

The maximum value is:

Maximum = 100 kg.

The first quartile is the median value of the first half of the data.

The first half of the data is:

S₁ = {45, 51, 53, 55, 55, 65, 75}

The median of this data set is the middle value, i.e. 4th value.

So, the  first quartile is:

Lower quartile = 55 kg.

The minimum value is:

Minimum = 45 kg.

The third quartile is the median value of the second half of the data.

The second half of the data is:

S₂ = {84, 87, 93, 93, 95, 96, 100}

The median of this data set is the middle value, i.e. 4th value.

So, the third quartile is:

Upper quartile = 93 kg.

The The median of the data set is the middle value, i.e. the 8th value.

Median = 81 kg.

6 0
3 years ago
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