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2 years ago

NEED HELPPPP PLEASEEE!!! NUMBER 5 ONLY

Mathematics

1 answer:

Reika [66]2 years ago

7 0

Answer:

Let x equal the ice thickness. An equality that represents a safe ice thickness for walkability is:

x ≥ 4 inches

(Plus the graph)

Step-by-step explanation:

Defining a variable just means you let any letter or symbol take the place of something. But you have to specifically say what is what in order for it to be clear.

So I defined "x" as the variable to represent the ice's thickness. And since we want an inequality for all the safe thicknesses, we could say that "x" must be greater than or equal to 4 inches thick in order to safely walk on it.

Lastly, you'd graph it with a solid point on 4 with the arrow going to the right.

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What is mZ1? 490 470 m21

zalisa [80]

490

Step-by-step explanation:

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2 years ago

A train leaves a station and travels north at a speed of 105105 km/h. twotwo hours later, a second train leaves on a parallel

hram777 [196]

Let x = hours to meet

train one has a 2 hour head start or 210 km head start

210 + 105x = 135x
30x = 210
x = 7 hours to meet

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3 years ago

If the circle below is cut from the square of plywood below, how many square inches of plywood would be left over?

Katyanochek1 [597]

Answer:

$= area \: of \: square - area \: of \: circle \\ = (s \times s) - (\pi {r}^{2} ) \\ = (24 \times 24) - (3.14 \times ( { \frac{24}{2} })^{2}) \\ \\ = 576 - 452.16 \\ = 123.84 \: square \: inches$

4 0

3 years ago

Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour

UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

$C(t) = C(0)e^{rt}$

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So $C(0) = 110$

Husband

Half life of 4 hours. So

$C(4) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{4r}$

$e^{4r} = 0.5$

Applying ln to both sides

$\ln{e^{4r}} = \ln{0.5}$

$4r = \ln{0.5}$

$r = \frac{\ln{0.5}}{4}$

$r = -0.1733$

So for the husband

$C(t) = 110e^{-0.1733t}$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.1733t}$

$C(11) = 110e^{-0.1733*11} = 16.35$

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

$C(10) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{10}$

$e^{10r} = 0.5$

Applying ln to both sides

$\ln{e^{10r}} = \ln{0.5}$

$10r = \ln{0.5}$

$r = \frac{\ln{0.5}}{10}$

$r = -0.0693$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.0693t}$

$C(11) = 110e^{-0.0693*11} = 51.33$

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0

3 years ago

Solve the system of equations below x-2y=3 2y-3y=9

Crank

First you solve 2y-3y=9.
$2y - 3y = 9 \\ subtract \: 2y \: and \: 3y \\ - y = 9 \\ divide \: by \: - 1 \\ y = - 9$
Now that you have y, then you substitute y into x-2y=3 to find X

$x - 2( - 9) = 3 \\ multiply \: - 9 \times - 2 \\ x + 18 = 3 \\ subtract \: 18 \: on \: both \: sides \\ x = - 15$
Your solution is (-15, -9)

6 0

3 years ago