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Maslowich
3 years ago
13

Please Help me with these question. first person to help me gets brainliest!!!!

Mathematics
1 answer:
lara [203]3 years ago
7 0

4) the distance between the x-values is 3 and the distance between the y-values is 4

3² + 4² = d²

9 + 16 = d²

25 = d²

√25 = d

5 = d

Answer: C

6)

d = \sqrt{(x2 - x1)^{2}  + (y2 - y1)^{2} }

d = \sqrt{(-2 - (-3))^{2}  + (2 - (-4))^{2} }

d = \sqrt{(-2 +3)^{2}  + (2 +4)^{2} }

d =  \sqrt{(1)^{2}  + (6)^{2} }

d =  \sqrt{1 + 36}

d = \sqrt{37}

d = 6.1

Answer: A

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Answer:

Sorry i don't know......!!!!!!!

Step-by-step explanation:

5 0
3 years ago
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If f(x) = 7(2-1)+8, what is the value of f(11) ?
andrew11 [14]

Answer:

f ( 2 ) = 20

Explanation:

To evaluate  f ( 2 )

substitute x = 2 into  f ( x )

f ( 2 ) = ( × 2 2 − ( 4 × 2 × x = 28− 8 = 20

8 0
2 years ago
100 grams/day of substance x is filtered, 50 grams/day is reabsorbed, 75 grams/day is secreted and 125 grams/day is excreted. su
Ivan

100 grams/day of substance x is filtered, 50 grams/day is reabsorbed, 75 grams/day is secreted and 125 grams/day is excreted. substance x is undergoing in the nephron net secretion and would have a clearance value that is greater than GFR

The formula for calculating clearance is Cx is given in terms of volume per time, where Ax is the quantity of x removed from the plasma, Px is the average plasma concentration, and Cx = Ax /Px.

Ax= 125 grams/day

and Px= 50 grams/day

Cx= 125/ 50

Cx= 2.5 grams/day

Therefore, the creatinine clearance overestimates GFR more severely in patients with a very low GFR than in those with higher GFR values. For a patient with a very low GFR, the secreted component is a relatively larger fraction of the total amount excreted.

Creatinine clearance remains the most popular method for routinely assessing patient GFR and the integrity of renal filtration, despite its low cost and convenience.

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Complete question :

Patient was experiencing respiratory acidosis (the increased hydrogen ion concentration was predominantly activating his peripheral chemoreceptors). 100 grams/day of substance X is filtered, 50 grams/day is reabsorbed, 75 grams/day is secreted and 125 grams/day is excreted. Substance X is undergoing ___________ in the nephron and would have a clearance value that is _________.(choose one)

A)net reabsorption, less than GFR

B)net secretion, greater than GFR

C)neither net reabsorption nor net secretion, equal to GFR

D)net reabsorption, greater than GFR

E) net secretion, less than GFR

3 0
1 year ago
A researcher found that the sea level rose 1 5 of a centimeter in 2 3 of a year. At what rate did the sea level rise? A) 2 5 cen
Ksenya-84 [330]
D) 3/10 per cen. year
4 0
3 years ago
Read 2 more answers
A professor gives her 100 students an exam; scores are normally distributed. The section has an average exam score of 80 with a
frozen [14]

Answer:

6.18% of the class has an exam score of A- or higher.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 80, \sigma = 6.5

What percentage of the class has an exam score of A- or higher (defined as at least 90)?

This is 1 subtracted by the pvalue of Z when X = 90. So

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

Z = \frac{90 - 80}{6.5}

Z = 1.54

Z = 1.54 has a pvalue of 0.9382

1 - 0.9382 = 0.0618

6.18% of the class has an exam score of A- or higher.

4 0
3 years ago
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