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3 years ago

Which of the following is the range thanks

Mathematics

1 answer:

const2013 [10]3 years ago

6 0

C. Its just the right hand column.:)

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Can someone please help?

djverab [1.8K]

Sure what do you need help on

6 0

3 years ago

Find the angle measures. Justify your responses. Given: a||b, m1=71º Find m5, m8

Natalka [10]

Answer:

The measure of angle MNP is

m ∠ MNP = (x - y)/2

Explanation:

The image attached shows the figure corresponding to this question.

The angle MNP, which is also the angle LNP, is formed by the intersection of a secant and a tangent to a circle.

Then, you can use the theorem:

the angle formed by a secant and a tangent to a circle that intersect outside the circle is half the difference of the major arc minus the minor arc.

The major arc formed is identified with the letter x and the minor arc is identified with the letter y. Thus, the measure of the angle MNP is half the difference x - y:

m ∠ MNP = (x - y)/2

Brainliest please

7 0

2 years ago

2x + x + x + 2y × 3y × y

Vilka [71]

Answer:

$\large\boxed{4x+6y^3}$

Step-by-step explanation:

$2x+x+x+2y\times3y\times y\\\\=(2x+x+x)+(2y\times3y\times y)\\\\=4x+(2\times3)(y\times y\times y)\\\\=4x+6y^3$

5 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

Triangle DEF has sides of length x, x+3, and x−1. What are all the possible types of DEF?

Natalija [7]

Triangle DEF is scalene

Must click thanks and mark brainliest

6 0

3 years ago