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mash [69]
3 years ago
10

Diego paid $47 for 3 tickets to a concert. Andre paid $141 for 9 tickets to a concert. Did they pay at the same rate?

Mathematics
2 answers:
frez [133]3 years ago
7 0

Answer:

They did pay at the same rate

Step-by-step explanation:

when dividing 47 by 3 you get 15.666... and when dividing 141 by 9 you also get 15.666... using this you can deduce that it is the same rate.

sweet-ann [11.9K]3 years ago
4 0

Answer:

Yes

Step-by-step explanation:

We can use ratios for this:

Diego:

47:3 tickets

Divide both sides by 3 so you get the price of one ticket

15.67: 1 ticket

Andre:

141: 9 tickets

Divide both sides by 9 so you get the price of one ticket

15.67: 1 ticket

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When buying bottles of Apple juice Ava spent $1.56 On a 12 -ounce bottle and $0.9 on a 6-ounce bottle. Use unit rates to determi
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Ok so you divide them and then make sure that 1 is the denominator and the numerator can be anything else
5 0
3 years ago
Chloe has enough sand to fill a sandbox with an area of 36 square units. She wants the outer edges of the sandbox to use as litt
tia_tia [17]

Answer:

  6 units by 6 units

Step-by-step explanation:

The quadrilateral with the smallest perimeter for a given area is the square.

The square with area 36 square units has side lengths of √36 = 6 units.

4 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
For the given expression, use only the Associative Property of Addition to enter an equivalent expression. 8 + a( +b ) =
Sergeeva-Olga [200]

Answer:

(8+a)+b

Step-by-step explanation:

8 0
3 years ago
Use the commutative property of multiplication to write a related number sentence 4×5=20
MrRissso [65]
Do you mean 4 x 5=20 turns into 5 x 4=20 or do you mean like 6 x 7=42
4 0
3 years ago
Read 2 more answers
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