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3 years ago

Guys please help with this question thanks

Mathematics

1 answer:

erastova [34]3 years ago

3 0

= 16 + 49 - 3(11) - 4(10)
= 16 + 49 - 33 - 40
= -8

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Jane and Sali cycled along the same 63km route. Jane took 3.5 hours to cycle the 63km.

Arisa [49]

Answer:

sali's speed was 18.75 km/h

Step-by-step explanation:

Jane took 3.5 hours to cycle the 63 km.

As, , so the speed of Jane will be:

Suppose, the speed of Sali is km/h

Sali caught up with Jane when they had both cycled 30 km.

So, the time required for Jane to cycle 30 km and the time required for Sali to cycle 30 km

Given that, Sali started to cycle 4 minutes or after Jane started to cycle. So, the equation will be.......

5 0

3 years ago

Which expression is equivalent to sec^2x − 1? a. cot2x b. tan2x c. csc2x d. cos2x

Nesterboy [21]

Sec^2 x - 1 = tan^2x

Proof:
Sec^2x = 1+ tan^2x

1/cos^2x = 1 + sin^2x/cos^2x

1/cos^2x - sin^2x/cos^2x = 1
Using common denominator:
(1-sin^2x)/cos^2x = 1
sin^2x + cos^2 x = 1
cos^2 x = 1 - sin^2x
Substituting :
cos^2x/cos^2x = 1
1 = 1
Left hand side = right hand side

4 0

3 years ago

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I need help with this problem if anyone can help ASAP please help !

MakcuM [25]

Answer:

Step-by-step explanation:

7 0

3 years ago

Consider the sets below.

otez555 [7]

B = CA

....hope this helps

3 0

3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago