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Flauer [41]
3 years ago
12

Express this as a difference: 7+3 Plz explain If you do you will get brainlyes

Mathematics
2 answers:
likoan [24]3 years ago
8 0
10 because if you 3 with 7 I will make 10




Did I blow your mind so hit that heart button
choli [55]3 years ago
6 0

Answer:

7-(-3)

Step-by-step explanation:

The answer is 7-(-3).

You're subtracting (-3) from 7.

7-(-3) = 7+3 because -(-3) = 3.

You might be interested in
TEN POINTS !!!!! if f(x)=x^2-6x and g(x)=x^3 squared what is (fog) (x)
Elena L [17]

Answer:

f(x) = x² - 6x

g(x) = (x³)² = x^6

( f o g) x

Wherever you see x in f(x) replace it by g(x)

That's

( f o g)(x) = (x^6)² - 6(x^6)

= x^12 - 6x^6

Hope this helps

5 0
3 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
What is the quotient?<br><br> -4/5 *2
kogti [31]

Answer:

-8/5

Step-by-step explanation:

convert the expression:-4/5 *2

multiply the fractions: -4/5 *2/1 (multiply across)

the you will get -8/5

4 0
3 years ago
Help me classify the triangles by its sides and measuring its angles :) ​
barxatty [35]

Answer:

Triangle 1 : isosceles triangle

Triangle 2 : Right Triangle

Step-by-step explanation:

5 0
3 years ago
What is the y-intercept of this graph?
melisa1 [442]

Answer:

The Y-intercept is 0.

Step-by-step explanation:

5 0
3 years ago
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