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2 years ago

A scuba diver swims 45 m below sea level. A bird flies 15 m above sea level. Write an inequality to represent this statement.

Mathematics

1 answer:

Wewaii [24]2 years ago

7 0

Answer: -45<15

Step-by-step explanation:

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Click the picture I need help please

Wittaler [7]

Answer:

Step-by-step explanation:

x=2

5*2+1=10+1=11

8 0

3 years ago

Read 2 more answers

when 55 is added to a certain number and the sum is divided by 3 , the result is 4 times the original number what is the origina

gogolik [260]

Answer:

4.99

Step-by-step explanation:

i cant remember how i got 4.99 but can show u how to check

= 55 +<u> 4.99 </u>= 59.99

59.99 divide <u>3</u> = 19.99666667

= 19.99666667 divide 4 (working in reverse as it says 4 times the orig no.)

= 4.99 (was rounded)

7 0

3 years ago

In what form is the following linear equation written?

krok68 [10]

Answer:

<h2>B. Standard</h2>

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The standard form of an equation of a line:

$Ax+By=C$

The general fom of an equation of a line:

$Ax+By+C$

We have the equation 3x - 2y = 4 in standard form.

5 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: