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Sveta_85 [38]
3 years ago
7

A white tailed deer can sprint at speeds up to 30 miles per hour America bison can run at speeds up to 3,520 feet per minute whi

ch animal is faster and by how many miles per hour? There are 5,280 feet in one mile
Mathematics
1 answer:
Tema [17]3 years ago
3 0

In 1 minute, bison runs 3520 feet

In 60 minutes, the bison would run

3520*60 feet

211200 feet per hour.

These are equivalent to;

40 miles per hour since 1 mile is equivalent to 5280 feet.

The bison is faster by 10 miles per hour

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4. A rectangular park has been constructed in
Helga [31]

Answer:

10ft

Step-by-step explanation:

Think of the rectangle as two right triangles. Sides a and b on the triangle are 8ft and 6 ft.  That's 8 squared + 6 squared = c squared.  C is 10 ft.

3 0
3 years ago
Help!! - 2.10 - (4 points)
Brrunno [24]

Answer:

Approach: Difference of Squares Pattern

4 {x}^{2}  - 25 = (2x - 5)(2x + 5)

Step-by-step explanation:

The given binomial is:

4 {x}^{2}  - 25

We can rewrite to obtain:

{(2x)}^{2}  -  {5}^{2}

This is a difference of two squares, so we will factor using difference of squares pattern.

Recall that:

{a}^{2} -  {b}^{2}   = (a + b)(a - )

If we let

a = 2x

and

b = 5

Then we can factor the given binomial to obtain:

{2x}^2 -  {5}^{2}  = (2x - 5)(2x + 5)

\therefore4 {x}^{2}  - 25 = (2x - 5)(2x + 5)

6 0
3 years ago
Which scale you would use on the coordinate plane to plot each set of points? A Select . (1.-6).(-7,-8), (-3, 7), (0.9) • (-20,
antoniya [11.8K]

Answer:qjs

Step-by-step explanation:

A

X= -7 to 1

Y= -8 to 9

B

X= -40 to 20

Y= -30 to 10

4 0
1 year 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
2 years ago
Can someone help me Please
alisha [4.7K]
Jane:  Think:  the hypotenuse of her ramp is 14 inches and the angle opposite the rise is 30 degrees.  Thus, cos 30 deg = (adj side) / (hyp) = (adj side) / (14 in).

Solving for (adj side), we get   (adj side) = (14 in)(sqrt(3)/2) = 7 sqrt(3) inches.
3 0
3 years ago
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