Can someone solve this?

I Need The Answer Plz!!! Geometry Is Hard!!!

Scorpion4ik [409]

Answer:

B. 10

Step-by-step explanation:

First, we need to recognize that the two triangles are similar (the triangle with a leg value of 21 and the triangle with a leg value of 4 and hypotenuse x)

Let's first label the leg of the second triangle as y.

Then, we can set up this equation

$\frac{21}{y} =\frac{y}{4}$

Cross multiplying gives us

84=y^2

So, y=sqrt(84) or 2 sqrt(21).

Now, using the pythagorean theorem, we get

4^2+2sqrt(21)^2=x^2

100=x^2

x=10

4 0

3 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

A boy purchased(bought) a party length sandwich 59 in long. He wants to cut it into three pieces so that the middle piece is 6in

Leviafan [203]

Ok so let's start with what we know- the shortest piece is 8 inches so there's one length... then the middle piece is 6 inches longer than the shortest (6+ 8) so the middle piece would be 14 inches long. To find the last piece we can add up the other two pieces we know (14+8) which would be 22 and subtract that from how long the whole sandwich is (59-22) which would be 37 inches long. So in the end he shortest piece would be 8 inches, the middle 14 inches and the longest 37 inches.

8 0

3 years ago

Simplify foundation of alegebra

Eva8 [605]

1.) -8 + 10(6 + 4r)

First multiply the 10 into (6 + 4r)

-8 + 60 + 40r

52 + 40r

2.) (8m² - 8m + 8) - (2m - 7m² - 6)

Distribute the - into (2m - 7m² - 6)

8m² - 8m + 8 - 2m + 7m² + 6

Combine like terms

15m² - 10m + 14

(idk if you need to factor this or if you can leave it as is)

3 0

4 years ago

In how many ways can a teacher arrange 5 students in the front row of a classroom with a total of 20 students?

aksik [14]

Slot method
5 slots

1st slot: 20
2nd slot: 19 (1 went to first slot)
3rd slot: 18
4th slot: 17
5th slot: 16

20*19*18*17*16=186,040 ways

5 0

3 years ago

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