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3 years ago

Length = 2 feet, width = 15 inches, height = 8 inches V= _____________

Mathematics

2 answers:

I am Lyosha [343]3 years ago

6 0

Volume = length * width ( aka base ) * height

2 feet = 24 inches.

24*15*8=2880 inches.

Gennadij [26K]3 years ago

4 0

V= 240 in. Hope this helps

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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

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

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3 years ago

Which graph represents 4x + 7y < -21

allsm [11]

Answer:

Step-by-step explanation:

Since the sign is not $\leq$ , and it is <, then you'd choose the option with the dotted line showing it isn't inclusive.

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

80°

Step-by-step explanation:

35° + 45° = 80°

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3 years ago

What is 6x^2-2x+1=0 plz help

Kazeer [188]

6x^2 - 2x + 1 is a quadratic formula from the form ax^2 + bx + c. This form of equation represents a parabola.
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.

In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.

I've added a picture of the parabola represented by this equation under the answer.

Hope this Helps! :)

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3 years ago