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3 years ago

What is the area of the triangle formed from (0,-1), (0.4), and (4,-1)?

Mathematics

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

10 squared units

Step-by-step explanation:

This triangle is pretty easy since it creates a right triangle. So the base length and height length are pretty easy to identify.

The area of a triangle is .5*base*height.

The base length after drawing the points is a horizontal so just count it or do 4-0=4.

The height length after drawing the points is a vertical so just count it or do 4-(-1)=4+1=5.

So the area of the triangle is .5*4*5=.5(20)=10 units squared.

Now if you didn't fill like drawing it and you knew how to find the determinant.

You just compute

.5 times det| 0 -1 1 |

| 0 4 1 |

| 4 -1 1 |

.5[ 0 det| 4 1 | - -1 det| 0 1 | + 1 det | 0 4 | ]

| -1 1 | | 4 1 | | 4 -1 |

.5[ 0 - -1(0-4) + 1(0-16)]

.5[ 0 -4 -16]

.5[-20]

-10

And if you get a negative just take the absolute value of it giving you 10.

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3 years ago

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

Hence the limit of the function $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

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Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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