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4 years ago

Can the sides of a triangle have lengths 6.2, 0.5, and 7.8? yes or no

1 answer:

4 0

No, it's not possible for the sides of a triangle to have those lengths.

According to the triangle inequality theorem, the sum of any two sides of the triangle has to be bigger than the last side. Let's test this.

$6.2+7.8\ \textgreater \ 0.5$

This inequality satisfies the triangle inequality theorem.

$7.8+0.5\ \textgreater \ 6.2$

This also satisfies the theorem.

$6.2+0.5 \textless \ 7.8$

Uh oh. This does not satisfy the triangle inequality theorem. Thus, it is not possible for a triangle to have these side lengths.

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Estimate each quotient tell <br> which method you used

cupoosta [38]

I would do 3+5 so then it equals 8

7 0

3 years ago

PLEASE I NEED HELP 10 points (dont waste them) What is the area of this figure?

meriva

Answer:

1280

Step-by-step explanation:

I think it is 1280 because 20*16*8/2=1280. I am not 100% sure.

4 0

3 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

3 years ago

skelet666 [1.2K]

Answer:

option D. 126 cm

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In this problem

Triangles PQR and XYZ are similar by AA Similarity Theorem

$\frac{XY}{PQ}=\frac{YZ}{QR}=\frac{XZ}{PR}$

Let

z ---> the scale factor

$z=\frac{XY}{PQ}$

substitute the given values

$z=\frac{30}{5}=6$

step 2

Find the perimeter of triangle XYZ

we know that

If two figures are similar, then the ratio of its perimeters is equal to the scale factor

Let

z ----> the scale factor

p_1 ----> the perimeter of triangle XYZ

p_2 ---> the perimeter of triangle PQR

$z=\frac{p_1}{p_2}$

The perimeter of triangle PQR is

$p_2=5+10+6=21\ cm$

we have

$z=6\\p_2=21\ cm$

substitute

$6=\frac{p_1}{21}$

$p_1=6(21)=126\ cm$

therefore

The perimeter of triangle XYZ is 126 cm

5 0

4 years ago

From the previous video: At age 40 you deposit $150 at the end of each month into an IRA that pays 4% interest

liberstina [14]

Answer: 32119.43

Step-by-step explanation:

3 0

3 years ago