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Mrrafil [7]
3 years ago
7

Rewrite the following equation in slope intercept form 14x+13y=12

Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0

Answer:

y = -\dfrac{14}{13}x + \dfrac{12}{13}

Step-by-step explanation:

The slope-intercept form of the equation of a line is

y = mx + b

To write the equation of a line in the slope-intercept form, solve the equation for y.

14x + 13y = 12

Subtract 14x from both sides.

13y = -14x + 12

Divide both sides by 13.

y = -\dfrac{14}{13}x + \dfrac{12}{13}

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vlabodo [156]

Answer:

m = 2, (0,4)

Step-by-step explanation:

Slope is 2

Y intercept is 0,4

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goblinko [34]
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Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream, it produces carbonic acid and lowers the pH. T
Irina18 [472]

Answer:

x ≅ 20.10 torr

Step-by-step explanation:

Given that the equation which models blood pH in the question is;

pH(x)=6.1+log(800/x)

where;

pH = 7.7

x = partial pressure of carbon dioxide in arterial blood, measured in torr.

we are asked to find (x)

In order to do that, we use the given equation:

pH(x)=6.1+log(800/x)

since pH = 7.7

7.7 = 6.1 + log (800/x)

7.7 - 6.1 =  log (800/x)

1.6 =  log (800/x)

10^{1.6}= \frac{800}{x}

x= \frac{800}{10^{1.6}}

x = 20.09509145

x ≅ 20.10 torr

4 0
3 years ago
Can u help me on this plzz
avanturin [10]
Median: 29
Range: 25
IQR: 14.5

Explanations:

**Median:**
(To find the median, we need to first order all the elements)

Ordered —> 15, 18, 18, 20, 23, 28, 30, 33, 33, 34, 38, 40

(Since there are an even number of elements, we need to add the two elements in the middle and divide by 2)

Median = (28 + 30)/2 = 58/2 = 29

**Range:**

(To find the range, you just have the subtract the smallest one from the largest)

Range = 40 - 15 = 25

**IQR:**

First half of elements —> 15, 18, 18, 20, 23, 28
Second half of elements —> 30, 33, 33, 34, 38, 40

Q1 (Quartile 1) = Median of first half = (18 + 20)/2 = 38/2 = 19
Q3 (Quartile 3) = Median of second half = (33 + 34)/2 = 67/2 = 33.5

IQR = Q3 - Q1 = 33.5 - 19 = 14.5
5 0
2 years ago
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