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3 years ago

PLS HELP ILL MARK U BRAINLIEST

Mathematics

1 answer:

vredina [299]3 years ago

3 0

Answer:

Step-by-step explanation:

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Which equation is not a linear equation?

Bond [772]

Answer:

2/x + 3/y = 5

Step-by-step explanation:

A linear equation refers to any equation that can be written in the form;

ax+b=0 or y - mx -c = 0

The graph of a linear equation will always be a straight line; Vertical, horizontal or with a given degree of steepness (slope).

All the equations given are linear, except the last equation;

2/x + 3/y = 5. The exponents of the variables x and y is not 1 hence the equation is non-linear. Find the graph attached;

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Help quick The engineers designing the All Aboard Railroad between Boca Raton and Jupiter decide to create parallel tracks throu

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Answer:

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Step-by-step explanation:

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3 years ago

Solve for x: -|x|>4

Feliz [49]

$-|x| > 4\ \ \ \ \ |\text{change the signs}\\\\|x| < -4\ FALSE!!!!$

We know |x| ≥ 0 for all real numbers. Therefore your answer is x ∈ ∅ (no solution).

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4 years ago

Find the volume of a cone if the diameter is 8 m and height is m. round the answer to the nearet tenth

Allisa [31]

Answer:

Step-by-step explanation:

diameter=8 m

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height h= ? m

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3 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

monitta

Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

7 0

3 years ago