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saw5 [17]
3 years ago
11

What is the problem to h-16<24=

Mathematics
1 answer:
ELEN [110]3 years ago
5 0
First, you have to add 16 to both sides because it's the opposite of subtracting.
h-16<24
  +16  +16
-16+16 will cancel each other out.
24+16=40
So, h<40
Or in other words, h is smaller than 40.

Hope it helps ^_^
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Can anyone help with this question?
maxonik [38]
Xia's flowers = may + june
xia's flowers = 5f + 8f
xia's flowers = 13f

the expression for xia's flower is 13f
5 0
3 years ago
For the problem -4(-5)(9), will the answer be positive or negative?
Scorpion4ik [409]

Answer:

Positive

Step-by-step explanation:

-4(-5)(9)

Negative cancels out negative, so it makes a positive:

20(9)

Simplify normally to get 180 as your final answer.

8 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
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uysha [10]

Answer:

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5 0
2 years ago
Line AC and line DB intersect at point P. Solve for angle BPQ.
taurus [48]

Answer:

Angle BPQ = 64°

Step-by-step explanation:

4x + 12 +2x = 90

6x + 12 = 90

     - 12    -12

6x = 78

x = 13°

BPQ = ((4(13) + 12)°

           (52 + 12)°

            64°

8 0
2 years ago
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