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Mandarinka [93]
3 years ago
14

A duck's mass at birth was 0.05 kilogram. The duck gained approximately 0.042 kilogram each week. After how many weeks is the du

ck's mass 0.890 kilogram?
Mathematics
2 answers:
N76 [4]3 years ago
7 0
0.890-0.05 = .840
.840/ 0.042 = 20
20 weeks

SashulF [63]3 years ago
4 0
21 weeks, I believe.
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AOC and BOD are diameters of a circle, centre O. Angles BAD and CDA are 90°. Prove that triangle AND and triangle DCA are congru
ololo11 [35]

Answer:

By using SAS which stands for Side, Angle, Side.

Step-by-step explanation:

This means that if we have two angles and side measures that are equal to the corresponding angles and side measures of the other triangle then they are congruent.

5 0
3 years ago
Help please explain how you got the answer
pychu [463]

Answer: V≈900 ft³

Step-by-step explanation:

Formula

V=πr²h

Given

r=5 ft

h=12 ft

Solve

V=πr²h

V=π(5)²(12)

V=π(25)(12)

V=300π

V≈300(3)

V≈900 ft³

Hope this helps!! :)

Please let me know if you have any questions

5 0
3 years ago
Helppp please I’ll give points to whoever answers it
sashaice [31]

Answer:

The solution is in the attached file

Step-by-step explanation:

Choose the examples you would like to use from the examples listed for both true and false

5 0
3 years ago
What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)
Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\

Now to solve the integral on the right hand side we shall use Integration by parts Taking 7t^{3} as first function thus we have

\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\

Again repeating the same procedure we get

=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\

Again repeating the same procedure we get

\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\

Now solving this integral we have

\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}

Thus we have

L[7t^{3}]=\frac{7\times 3\times 2}{s^4}

where s is any complex parameter

5 0
3 years ago
Please help me I need this fast
Minchanka [31]
100 degrees I think....

I’m writing this because it said it was too short
4 0
3 years ago
Read 2 more answers
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