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3 years ago

5x + 13-14 or-4x + 6< 10

Mathematics

1 answer:

Blizzard [7]3 years ago

7 0

Answer:

5x - 1 or x>−1

Step-by-step explanation:

Just combine like terms: 13,-14

Combine like terms: 6,10

Divide by -4:-4x/-4<4/-4

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Let A = {1, 2, 3, 4, 5} and B = {2, 4}. What is A ∩ B?

Harlamova29_29 [7]

$A\cap B=\{x:x\in A \wedge x\in B\}$

$A\cap B=\{2,4\}$

3 0

3 years ago

Read 2 more answers

Pls answer this is due in 30 minutes

ziro4ka [17]

Answer:

1a) t = 0.17b

1b) $36.03

2) 51,167.50

Step-by-step explanation:

1a) 17% converted to a decimal (divide by 100) = 0.17. Multiply that by the amount of the bill (before tipping) to get the amount of the tip

1b) The bill of $248 represents the entire bill (100%) plus the tip amount (17%) which makes 117%. As a decimal, this is 1.17.

To find what the cost was without the tip, divide the bill total by 1.17

248/1.17 = 211.97. That was the bill before the tip was added.

248-211.97 = 36.03 was the tip added.

2) 5.5% converted to a decimal is 0.055

0.55 x 48,500 = 2667.50

$2,667.50 + $48,500 = 51,167.50

Or, you can use what we did in 1b, and his new salary is 1.055 times 48,500

1.055 x 48,500 = 51,167.50

6 0

3 years ago

Simplify the ratio 24:81

gizmo_the_mogwai [7]

Answer:

8 to 27

Step-by-step explanation:

3 0

3 years ago

Show work/pls explain

Tamiku [17]

9+6(2^2+4)
9+6(4+4)
9+(6)(8)
9+48
=57

5 0

3 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

3 years ago