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Jobisdone [24]
3 years ago
13

Indices-find The value of 6*7^3​

Mathematics
1 answer:
artcher [175]3 years ago
5 0

Answer:

6 * 7^3 = 2058

Step-by-step explanation:

Given

6 * 7^3

Required

Solve

App;y law of indices:

6 * 7^3 = 6 * 7*7*7

Multiply

6 * 7^3 = 2058

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Of 350 students polled, some have a cat, some have a dog, and some have both. One hundred seventy-five students have a cat and 1
BigorU [14]
Total students 350
350- 175 cats= 175
175 - 140 both pets = 35 dog only students

I guess
6 0
4 years ago
Find \(\int \dfrac{x}{\sqrt{1-x^4}}\) Please, help
ki77a [65]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2867785

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}


Make a trigonometric substitution:

\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}


so the integral (i) becomes

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}

\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}


Now, substitute back for t = arcsin(x²), and you finally get the result:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}          ✔

________


You could also make

x² = cos t

and you would get this expression for the integral:

\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}          ✔


which is fine, because those two functions have the same derivative, as the difference between them is a constant:

\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}

\mathsf{=\dfrac{\pi}{4}}         ✔


and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.


I hope this helps. =)

6 0
3 years ago
Find the solution of y = 5x – 4 for x = –6.
spayn [35]
The answer is (-6,-34)
Because since x = -6 u substituted so the equation will be y= 5(-6)-4 so y=-34 making the answer (-6,-34)
4 0
3 years ago
75 is what percent of 1000?
sineoko [7]
75 =   \frac{x}{100} \times 1000\\75 =x \times 10\\x =7.5 so your answer is 7.5%
3 0
3 years ago
Read 2 more answers
You have a new pool and want to know it's volume. The pool is 5 feet deep and has a radius of 7 feet. About how much water can t
MrRa [10]

Answer:

769 cubic ft

Step-by-step explanation:

Its volume:

Find the volume of a cylinder of radius 7 ft and length 5 ft:

The basic formula for his volume is V = πr²h.  In this case, the volume is:

V = 3.14(7 ft)²(5 ft) = 769 cubic feet of water.

3 0
3 years ago
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