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4 years ago

What is the equation has intercepts at x(1, 0, 0), y(0, -1,0), and z(0, 0, 2)

Mathematics

1 answer:

bulgar [2K]4 years ago

3 0

In intercept form, the plane that has these intercepts is ...

... x/(x-intercept) + y/(y-intercept) + z/(z-intercept) = 1

... x/1 + y/(-1) + z/2 = 1

... 2x -2y +z = 2 . . . . . in standard form

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Amir bought a camera 5 year ago for RM1500. He plans to sell his camera at a price of RM1800. What percentage of the profit or l

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3 years ago

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What is the value of A? -3x -3x² -5x –5x²

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3 years ago

Write 42 as a product of primes.

faltersainse [42]

Answer:

The prime factorization of 42 is 2 × 3 × 7

Step-by-step explanation:

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3 years ago

Please help me with this

drek231 [11]

Answer:

100 in²

Step-by-step explanation:

Since the figures are similar

the linear ratio of sides = a : b , then

ratio of areas = a² : b²

ratio of sides = 15 : 21 = 5 : 7

ratio of areas = 5² : 7² = 25 : 49

let the area of the smaller figure be x then by proportion

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3 years ago

Lim 1 - cos 40 x>01 - cos 60

Nuetrik [128]

Answer:

The answer is 4/9 if the problem is:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$ .

Step-by-step explanation:

I think this says:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$ .

Please correct me if I'm wrong about the problem.

Here are some useful limits we might use:

$\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1$

$\limg_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0$

So for our limit... I'm going to multiply top and bottom by the conjugate of the bottom; that is I'm going to multiply top and bottom by $1+\cos(6\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}\cdot\frac{1+\cos(6\theta)}{1+\cos(6\theta)}$

When you multiply conjugates you only have to do first and last of FOIL:

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{1-\cos^2(6\theta)}$

By the Pythagorean Identities, the denominator is equal to $\sin^2(6\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{\sin^2(6\theta)}$

I'm going to divide top and bottom by $36\theta^2$ in hopes to use the useful limits I mentioned:

$\lim_{\theta \rightarrow 0}\frac{\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{36\theta^2}}{\frac{\sin^2(6\theta)}{36\theta^2}}$

Let's tweak our useful limits I mentioned so it is more clear what I'm going to do in the following steps:

$\lim_{\theta \rightarrow 0}\frac{\sin(6\theta)}{6\theta}=1$

$\lim_{\theta \rightarrow 0}\frac{\cos(4\theta)-1}{4\theta}=0$

The bottom goes to 1. The limit will go to whatever the top equals if the top limit exists.

So let's look at the top in hopes it goes to a number:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot (1+\cos(6\theta)}$

We are going to multiple the first factor by the conjugate of the top; that is we are multiply top and bottom by $1+\cos(4\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot \frac{1+\cos(4\theta)}{1+\cos(4\theta)} \cdot (1+\cos(6\theta)}$

Recall the thing I said about multiplying conjugates:

$\lim_{\theta \rightarrow 0}\frac{1-\cos^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

We are going to apply the Pythagorean Identities here:

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{\frac{9}{4}(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{4}{9}\frac{\sin^2(4\theta)}{(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

Ok this looks good, we are going to apply the useful limits I mentioned along with substitution to find the remaining limits:

$\frac{4}{9}(1)^2 \frac{1+\cos(6(0))}{1+\cos(4(0))}$

$\frac{4}{9}(1)\frac{1+1}{1+1}$

$\frac{4}{9}(1)\frac{2}{2}$

$\frac{4}{9}(1)$

$\frac{4}{9}$

The limit is 4/9.

8 0

3 years ago