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3 years ago

What is the 98+56-52%=

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

5 0

Hopes this helps if u have any questions pls ask!!

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Evaluate triple integral

kaheart [24]

Answer:

$\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}$

Step-by-step explanation:

$\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\$

$\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\$

$\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\$

$\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\$

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

6 0

2 years ago

-21 +3 = -9r in the form of -2(x-p)2+q=

denis-greek [22]

Answer:

1.28 Step-by-step explanation: got

4 0

2 years ago

Our class has 50 students. Assuming that no students are born on leap days (February29), what is the probability that no two stu

madreJ [45]

Answer: a) $\dfrac{799}{31250}$ b) $\dfrac{30451}{31250}$

Step-by-step explanation:

Since we have given that

Number of days in a year = 365

Number of students = 50

Probability that no two students share the same birthday. It means each student has different birthday.

Probability(no two students share the same birthday) $=\dfrac{365-1}{365}\times \dfrac{365-2}{365}........\times \dfrac{365-50}{365}$

Probability(no two students share the same birthday) $=\dfrac{^{365}P_{50}}{365^{50}}$

Probability(no two students share the same birthday) $=\dfrac{799}{31250}\approx 0.025568$

Probability that at least one of the students has the same birthday as another student in the class is given by

$1-P(\text{no student share the same birthday})\\\\=1-\dfrac{799}{31250}\\\\=\dfrac{30451}{31250}=0.974432$

5 0

3 years ago

The ratio of the circumference of two circles is 3:2.What is the ratio of their area

Nezavi [6.7K]

The ratio of their area is 9/4

<h2>Explanation:</h2>

We use ratios to compares values. In this exercise, we are comparing the circumference of two circles, which is:

$3:2$

and we want to know what is the ratio of their area. Recall that the circumference of a circle is given by:

$C=2\pi r \\ \\ \\ Where: \\ \\ C:Circumference \\ \\ r:Radius \ of \ the \ circle$

If we define:

$C_{1}: \ Circumference \ of \ circle \ 1 \\ \\ C_{2}: \ Circumference \ of \ circle \ 2 \\ \\ r_{1}: \ Radius \ of \ circle \ 1 \\ \\ r_{2}: \ Radius \ of \ circle \ 2$

Then, the ratio of the circumference of two circles is 3:2 is:

$\frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{3}{2} \\ \\ \therefore \frac{r_{1}}{r_{2}}=\frac{3}{2}$

The area of a circle is given by:

$A=\pi r^2 \\ \\ A:Area \\ \\ r:Radius$

So the ratio of their area can be found as:

$\frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^2}{\pi r_{2}^2} \\ \\ \\ A_{1}:Area \ of \ circle \ 1 \\ \\ A_{2}:Area \ of \ circle \ 2$

So:

$\frac{A_{1}}{A_{2}}=\frac{r_{1}^2}{r_{2}^2}=\left( \frac{r_{1}}{r_{2}} \right)^2 \\ \\ \frac{A_{1}}{A_{2}}=\left(\frac{3}{2}\right)^2 \\ \\ \boxed{\frac{A_{1}}{A_{2}}=\frac{9}{4}}$

<h2>Learn more:</h2>

Unit rate: brainly.com/question/13771948#

#LearnWithBrainly

5 0

3 years ago

Is three and a half bigger or smaller than three and four tenth??

TiliK225 [7]

3 and 1/2 is bigger than 3 and 4/10

7 0

3 years ago