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adell [148]
3 years ago
7

Find the distance and midpoint for the points (5,5) and (1,2).

Mathematics
1 answer:
nikklg [1K]3 years ago
4 0

Answer: wtfffff

Step-by-step explanation:

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How do you do domain and range​
Agata [3.3K]

Answer:

the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis

4 0
3 years ago
A cylinder with radius 7 and height 4 is in table,then find its volume?<br>​
Lady bird [3.3K]

Answer:

V = ≈615.75

Step-by-step explanation:

hopefully that helped

8 0
3 years ago
PLEASE HELP ME
asambeis [7]
For the equation y - x < 0,
y < x
The shaded portion will be under the dashed line joining points (-2, -2) and (3, 3).

for the equation x - 1 > 0,
x > 1
The shaded portion will be to the right of the dashed line joining points (1, 5) and (1, -3).

The portion affected by the two shadings will be the portion common to the right of the dashed line joining points (1, 5) and (1, -3) and under the dashed line joining points (-2, -2) and (3, 3).
Therefore the correct answer is option c.
3 0
3 years ago
Read 2 more answers
The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by K(x) 5x/x^2
Olegator [25]

Given :

The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by K(x) = \dfrac{5x}{x^2+9}.

To Find :

Find the time at which the concentration is a maximum. b. Find the maximum concentration.

Solution :

For maximum value of x, K'(x) = 0.

K'(x) = \dfrac{5(x^2+9)- 5x(2x)}{(x^2+9)^2}=0\\\\5x^2+45-10x^2=0\\\\5x^2 = 45\\\\x = \pm 3

Since, time cannot be negative, so ignoring x = -3 .

Putting value of x = 3, we get, K(3) = 15/( 9 + 9) = 5/6

Therefore, maximum value drug in bloodstream is 5/6 at time x = 3 units.

Hence, this is the required solution.

5 0
2 years ago
Find the sum.
hichkok12 [17]

Answer:

C - 19/24

Step-by-step explanation:

Take the fractions and put the denominators into a common factor

\frac{1}{3}+ \frac{1}{3}+ \frac{1}{8}

\frac{8}{24}+ \frac{8}{24}+ \frac{3}{24}

Add Numerators Across

\frac{16}{24}+ \frac{3}{24}

\frac{19}{24}

8 0
3 years ago
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