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3 years ago

Maureen is taking an antibiotic. The table below shows the amount of antibiotic

Mathematics

1 answer:

Tems11 [23]3 years ago

7 0

Answer:

Answer: Maureen's body eliminated the bacteria faster then Ken's as if you multiply 150 times 0.976 you get a number larger then 90 so Ken would have eliminated 150 mg of antibiotics slower then Maureen ;We will use the equation for Ken on Maureen.

Step-by-step explanation:

f(t) =

So, in the first hour,

f(t) = = 195.2 mg

In the second hour,

f(t) = = 190.52 mg

In the third hour,

f(t) = = 186 mg

Now, comparing these values with Maureen's, we can see that she has a faster rate of 150,90,54 in the 1st, 2nd and 3rd hours respectively as compared to Ken.

So, we can conclude that Maureen's body eliminated the antibiotic faster than Ken's body.

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Drag the tiles to the correct boxes to complete the pairs. Find the distance between each pair of points. 5 units 4 units 2 unit

Olin [163]

Answer:

a) Distance between points A (5, 4) and B( 5, -2) is 6 units

b) Distance between points E (-2, -1) and F( -2, -5) is 4 units

c) Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) Distance between points G(3, -5) and H(6, -5) is 3 units

Step-by-step explanation:

We need to find the distance between each pair

a) A (5, 4) and B( 5, -2)

the distance formula is:

$d(A,B) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Putting values: x₁ = 5, x₂=5 and y₁= 4 and y₂= -2

$d(A,B) = \sqrt{(5-5)^2+(-2-4)^2}$

$d(A,B) = \sqrt{(0)^2+(-6)^2}$

$d(A,B) = \sqrt{36}$

$d(A,B) = 6$

Distance between points A (5, 4) and B( 5, -2) is 6 units

b) E (-2, -1) and F( -2, -5)

the distance formula is:

$d(E,F) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Putting values: x₁ = -2, x₂=-2 and y₁= -1 and y₂= -5

$d(E,F) = \sqrt{(-2-(-2))^2+(-5-(-1))^2}$

$d(E,F) = \sqrt{(0)^2+(-4)^2}$

$d(E,F) = \sqrt{16}$

$d(E,F) = 4$

Distance between points E (-2, -1) and F( -2, -5) is 4 units

c) C (-4, 1) and D( 1, 1)

the distance formula is:

$d(C,D) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Putting values: x₁ = -4, x₂=1 and y₁= 1 and y₂= 1

$d(C,D)= \sqrt{(1-(-4))^2+(1-(1))^2}$

$d(C,D) = \sqrt{(5)^2+(0)^2}$

$d(C,D) = \sqrt{25}$

$d(C,D) = 5$

Distance between points C (-4, 1) and D( 1, 1) is 5 units

d) G(3, -5) and H(6, -5)

the distance formula is:

$d(G,H) = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Putting values: x₁ = 3, x₂=6 and y₁= -5 and y₂= -5

$d(G,H)= \sqrt{(6-(3))^2+(-5-(-5))^2}$

$d(G,H) = \sqrt{(3)^2+(0)^2}$

$d(G,H) = \sqrt{9}$

$d(G,H) = 3$

Distance between points G(3, -5) and H(6, -5) is 3 units

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3 years ago

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STatiana [176]

Answer:

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Step-by-step explanation:

Put the values of x and y in the equations and solve the resulting system.

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The first equation tells the value of A:

2A = 8 . . . . add 3

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This and the second equation tells the value of B:

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Answer:

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Step-by-step explanation:

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3 years ago