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Anarel [89]
3 years ago
13

This is 5th grade math [ 1,225 - ( 568 - 203 )] × 10 [ 1,225 - ______ ] × 10 ________ × 10 ____

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
5 0
[1,225 - (568 - 203)] x 10
[1,225 - 365] x 10
860 x 10
8,600
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Someone please help me! I would appreciate it so much!
Alexxandr [17]
5.1 • 0.79 = 4.029
the apples cost $4.03
8 0
3 years ago
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PLEASE HELP: What is the distance in altitude between the top of a 5000 ft mountain and the bottom of a 28 foot well?
dedylja [7]
5028ft distance in the altitude between the top and bottom
5 0
3 years ago
The graph below describes the journey of a train between two cities
UkoKoshka [18]

Answer:

Step-by-step explanation:

To find : Acceleration in first 15 min . Distance between two cities  Average speed of journey

Solution:

Each horizontal block is 1/8 hr = 7.5 min

Each vertical block is 10 km/hr

Time                    Velocity  km/hr

0 Min  ( 0 hr)            0

15 Min (1/4 hr)           50

45 Min (3/4 hr)         50

60 MIn  ( 1 hr)            100

90 Min  ( 3/2 hr)        100

120 Min ( 2hr)            0

Acceleration in first 15 min  (1/4 hr)  =  (50 - 0)/(1/4 - 0)  = 50/(1/4)

= 200  km/h²

Distance between two cities

= (1/2)(0 + 50)(1/4 - 0)  + 50 * (3/4 - 1/4)  + (1/2)(50 + 100)(1 - 3/4)  + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)

=  25/4  + 25 + 75/4   + 50 + 25

= 125

Distance between two cities = 125 km

Average Speed of journey = 125/2  = 62.5  km/hr

Acceleration in first 15 min = 200  km/h²

Distance between two cities = 125 km

Average Speed of journey =  62.5  km/hr

Hope this helps..

3 0
3 years ago
Read 2 more answers
1.The graph of g is a translation 4 units up and 5 units right of the function
Naddika [18.5K]

Answer:

g(x) = 5(x - 5) + 4

Step-by-step explanation:

None of the answers listed are correct

4 0
2 years ago
Integrate sin^-1(x) dx<br><br> please explain how to do it aswell ...?
Lynna [10]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2264253

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}


Trigonometric substitution:

\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}


then,

\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}


So the integral \mathsf{(ii)} becomes

\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}


Integrate \mathsf{(ii)} by parts:

\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}

\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}


Substitute back for the variable x, and you get

\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>

6 0
3 years ago
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