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Marianna [84]
3 years ago
7

Yolanda wants to pour 91.48 grams of salt into a container so far she has poured 72.2 grams how much more salt should Yolanda po

ur
Mathematics
2 answers:
kvv77 [185]3 years ago
8 0
91.48-72.2= 19.28 grams= the answer
NeTakaya3 years ago
5 0

Answer:

19.28

Step-by-step explanation:

91.48 minus 72.2

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14 divided by 3 (27-11) x 3
stira [4]

14 ÷3 (27-11) x 3 = 224

4 0
4 years ago
The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given
slavikrds [6]

Answer:

a) \frac{dm}{dt} = -k\cdot m, b) m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }, c) m(t) = 10\cdot e^{-\frac{t}{2438.155} }, d) m(300) \approx 8.842\,g

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

\frac{dm}{dt} = -k\cdot m

b) The general solution is found after separating variables and integrating each sides:

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

Where \tau is the time constant and k = \frac{1}{\tau}

c) The time constant is:

\tau = \frac{1690\,yr}{\ln 2}

\tau = 2438.155\,yr

The particular solution of the differential equation is:

m(t) = 10\cdot e^{-\frac{t}{2438.155} }

d) The amount of radium after 300 years is:

m(300) \approx 8.842\,g

4 0
3 years ago
Read 2 more answers
The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate
soldi70 [24.7K]

Answer:

a) V = 33510.322\,in^{3}, b) A_{s} = 5026.548\,in^{2}, c) \% V = 0.450\,\%, \%A_{s} = 0.300\,\%.

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

V = \frac{4}{3}\pi \cdot r^{3}

A_{s} = 4\pi \cdot r^{2}

a) The volume of the sphere is:

V = \frac{4}{3}\pi \cdot (20\,in)^{3}

V = 33510.322\,in^{3}

b) The surface area of the sphere is:

A_{s} = 4\pi \cdot (20\,in)^{2}

A_{s} = 5026.548\,in^{2}

c) The total differentials for volume and surface area of the sphere are, respectively:

\Delta V = 4\pi\cdot r^{2}\,\Delta r

\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)

\Delta V = 150.796\,in^{3}

\Delta A_{s} = 8\pi\cdot r \,\Delta r

\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)

\Delta A_{s} = 15.080\,in^{2}

Relative errors are presented hereafter:

\%V = \frac{\Delta V}{V}\times 100\%

\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%

\% V = 0.450\,\%

\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%

\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%

\%A_{s} = 0.300\,\%

4 0
3 years ago
('h(I) = -35<br>(h(n) = h(n - 1)×2<br>h(3) =​
qwelly [4]

Answer:

h(3)=n(h_3)×3that is the answer

3 0
3 years ago
Solve this equation: 4y + 228 = 352.
Rzqust [24]
4y + 228= 352
      -228    -228
4y= 124
----   -----
4        4
y=31
5 0
3 years ago
Read 2 more answers
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