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3 years ago

Find the rate of change of the function h(x) = 2 x on the interval 2 ≤ x ≤ 4. The rate of change is a0.'

Mathematics

2 answers:

zheka24 [161]3 years ago

8 0

Answer:

Step-by-step explanation:

The rate of change of a linear function is the slope m

h(x) = 2x is a linear function with slope m = 2

Hence rate of change is 2

IceJOKER [234]3 years ago

7 0

Answer:

Step-by-step explanation:

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21 x ________ ÷ 3 = 21 *

GrogVix [38]

Answer:

first one: 3

second one: b

third one: 3.2

Step-by-step explanation:

first one:

21 x ________ ÷ 3 = 21

rewrite equation:

21 ÷ 3 x ___ = 21

7 x ___ = 21

(21 / 7 = 3)

7 x 3 = 21

second one:

b x h ÷ h = ___

third one:

876 x 3.2 ÷ ______ = 876

2803.2 ÷ ___ = 876

(2803.2 / 876 = 3.2)

2803.2 x 3.2 = 876

7 0

3 years ago

Which of the following equations is equivalent to log(y)= 3.994 ?

Kisachek [45]

Answer:

The correct answer is option D: $y = 10^{3.994}$

Step-by-step explanation:

Given:

log(y)= 3.994

Solution:

A logarithm base b of a positive number x satisfies the following definition:

$y =log_b(x) => b^y =x$

For $x > 0, b >0, b \neq 1$

Also if no base b is indicated, the base of the logarithm is assumed to be 10 .

Thus, in log(y)= 3.994 base b is not indicated. so its base is assumed to be 10

now

$log_{10}(y)= 3.994$

Then

$y = 10^{3.994}$

4 0

3 years ago

You are looking at a convention bid that requires 100,000 square feet of meeting space. You have one room at 200 ft x 200 ft, an

Alla [95]

The 2 rooms only add up to 45,000 square feet (150 x 150 x 2) and the larger room is only 40,000 square feet (200 x 200) so this only gives you 85,000 square feet (40,000 + 45,000) which is not enough for the 100,000.

7 0

3 years ago

16. If the base of parallelogram is 10 cm and its corresponding height is 5

lara [203]

Answer:

50 cm²

The answer is that because the area of a parallelogram is height × width.

height = 10cm

width = 5 cm

10 × 5= 50cm²

7 0

2 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

4 0

2 years ago