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2 years ago

Identify whether each cube root or square root lies between 2 and 3, 4 and 5, or 9 and 10.

Mathematics

1 answer:

leva [86]2 years ago

3 0

Answer:

The cube root and square root are not given, but I'm going to give a chart that might help.

Square roots:

2 = √4

3 = √9

4 = √16

5 = √25

9 = √81

10 = √100

Cube roots:

2 = ^3√8

3 = ^3√27

4 = ^3√64

5 = ^3√125

9 = ^3√729

10 = ^3√1000

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Eriko cuts out a paper quadrilateral with two angles that each measure 60° and two angles that each measure 120°. Only two of th

Whitepunk [10]

Answer:

The quadrilateral is Parallelogram.

Step-by-step explanation:

A quadrilateral is a four-sided regular polygon.

A parallelogram is a quadrilateral.

The opposite sides of a parallelogram are parallel and equal in length.

The opposite angles of a parallelogram are also equal.

In this case the quadrilateral is defined as follows:

two angles that each measure 60°
two angles that each measure 120°
Only two of the sides have a length of 3.2 inches.

Consider the quadrilateral ABCD below.

The quadrilateral ABCD is Parallelogram.

8 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

Write the inequality for the graph

Svetach [21]

Answer:

x≤-2

Step-by-step explanation:

it would be that because if the graph is going to the left that means it is less than x and the opposite is for greater than. Now if there is a closed circle that means the sign is equal to hence the line under the sign. So therefore it's x≤-2

5 0

3 years ago

2 POS

Lostsunrise [7]

Answer:

A. x=0

Step-by-step explanation:

x+20+10x=20+9x

cancel equal terms

x+10x+9x

add x to 10x

11x=9x

move variable to left

11x-9x=0

subtract 11x to 9x

2x=0

divide Both sides by 2

x=0

4 0

2 years ago

find the value of x2 values that separate the middle 90% from the rest of the distribution for 8 degrees of freedom

Anettt [7]

Answer:

With alpha 0.95 and 8 degrees of freedom χ²= 2.73

And with alpha 0.05 and 8 degrees of freedom χ²=15.51

Step-by-step explanation:

The significance level ∝ = 1- 0.9 = 0.1

But we need the area of the middle so we divide this significance level with 2

so that we get exactly the middle area .

Dividing 0.1/2= 0.05

So we will have two values for chi square

One with 0.9 + 0.05 = 0.95 alpha and one with 0.05 alpha . This is because the chi square is right tailed.

So with alpha 0.95 and 8 degrees of freedom χ²= 2.73

And with alpha 0.05 and 8 degrees of freedom χ²=15.51

This can be shown with a graph.

4 0

3 years ago