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Serhud [2]
3 years ago
6

The temperature in Sheffield is -7c

Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0

Answer:

24c

Step-by-step explanation:

Difference= big value - small value

When minus is put side by side, it becomes a plus.

17c-(-7c)=17c+7c=24c

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URGENT HELP NEEDED!!!!!
Alona [7]

Answer:

k

Step-by-step explanation:

7 0
3 years ago
An oblique triangular pyramid has a base area
lubasha [3.4K]

The height of the oblique triangular pyramid is 18 units if the base area of 21 square units and a volume of 126 cubic units.

<h3>What is volume?</h3>

It is defined as a three-dimensional space enclosed by an object or thing.

We know the volume of an oblique triangular pyramid is given by:

\rm V = \dfrac{1}{3}b\times h

Where b is the area of the base and h is the height of the pyramid.

We have b = 21 square units  and V =  126 cubic units.

\rm 126 = \dfrac{1}{3}\times21\times h

h = 18 units

Thus, the height of the oblique triangular pyramid is 18 units if the base area of 21 square units and a volume of 126 cubic units.

Learn more about the volume here:

brainly.com/question/16788902

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5 0
2 years ago
Irina ran 0.25 mile in 2 minutes. At this rate, how many minutes will it take her to run 2 miles?
scoray [572]

Answer:

16

Step-by-step explanation:

 2/0.25 = 8

2 x 8 = 16

4 0
3 years ago
Read 2 more answers
Tell whether the two rates form a proportion:<br> 35 cars in 5 days<br> 60 cars in 10 days
monitta
Yes they both are proportional. you can do 7 cars each day for 5 days. and then you can do 6 cars a day for 10 days
6 0
3 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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