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2 years ago

Divide;11dq^4/8d^3q^4÷5d/10d^2q^3

Mathematics

1 answer:

Blababa [14]2 years ago

5 0

Answer: 664546^74767^-6382639r575

Step-by-step explanation:

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6 points are placed on the line a , 4 points are placed on the line b . How many triangles is it possible to form such that thei

BaLLatris [955]

Answer:

Step-by-step explanation:

Using side b as the base, 4 points makes 3 bases (the space in between). With three bases, you can have 3 bases of 1 segment, 2 bases of 2 segments, and 1 base of 3 segments. This equals 6 bases. Each of these can connect to a point on line a. 6x6=36

Using side a as the base, 6 points makes 5 bases. With 5 bases, you can have 5 bases of 1 segment, 4 bases of 2 segments, 3 bases of 3 segments, 2 bases of 4 segments, and 1 base of 5 segments. This equals 15 bases. Each of these can connect to a point on line b. 15x4=60

36+60=96

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2 years ago

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30+5y_>4x I need the answer ASAP

Gre4nikov [31]

30+5y_>4x
5y_>4x-30
The answer is: y_>(4/5)x-6

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2 years ago

Find x of the triangle

sukhopar [10]

Answer:

Step-by-step explanation:

altitude = sqrt(6*8) = 4*sqrt(3)

Use the Pythagorean Theorem to derive x

altitude^2 + 18^2 = x^2

48 + 18^2 = x^2

x^2 = 342

x = 3*sqrt(38)

x = 18.49

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2 years ago

What information does a supply schedule provide?

dimulka [17.4K]

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Step-by-step explanation:

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2 years ago

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

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, 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