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3 years ago

Graph the line with 1/2 slope passing through the point (-5,1).

Mathematics

1 answer:

Ghella [55]3 years ago

5 0

Answer:

Go on dezmos it will do it for you if you just wright the coordinates in

Step-by-step explanation:

You might be interested in

The base of a hexagonal prism has an area of 96 square inches. The height of the prism is 9 inches. Find the volume of the prism

Kitty [74]

Answer:

The volume of the prism is 864 inches^3.

Step-by-step explanation:

The basic formula for the volume of a prism is:

V = Ah

where:

V = Volume of prism

A = Area of uniform cross-section or base of prism = 96 inches^2

h = Perpendicular height of prism = 9 inches

Substitute the values given in the question into the formula.

V = 96 × 9

V = 864 inches^3

7 0

4 years ago

Select ALL the correct answers.

shepuryov [24]

When a quadrilateral is inscribed in a circle, the opposite angles are supplementary

The description of the angles in the quadrilaterals are:

b. m∠M = 55°, m∠J = 48°, and m∠L = 132°
d. m∠L = 40°, m∠M = 60°, and m∠K = 120°
e. m∠K = 72°, m∠L = 44°, and m∠M = 108°
f. m∠J = 105°, m∠K = 65°, and m∠L = 75°

<h3>How to describe the angles</h3>

The quadrilateral is given as: JKLM

The opposite angles are:

Angles J and L
Angles K and M

The opposite angles are supplementary

So, we have:

$J + L = 180$

$K + M = 180$

Next, we test the options

Option (a)

$J + L = 180$

$125 + 35 = 160$

This is not true

Option (b)

$J + L = 180$

$48 + 132 =180$

This is true

Option (c)

$K + M = 180$

$12 + 78 = 180$

This is not true

Option (d)

$K + M = 180$

$120 + 60 = 180$

This is true

Option (e)

$K + M = 180$

$72 + 108= 180$

This is true

Option (f)

$J + L = 180$

$105 + 75 = 180$

This true

Hence, the description of the angles in the quadrilaterals are (b), (d), (e) and (f)

Read more about inscribed quadrilaterals at:

brainly.com/question/26690979

6 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

4 years ago

Helpp meee pleaseeee

ludmilkaskok [199]

Answer:

I think its b:/ srry if I'm wrong

5 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

Alenkasestr [34]

Answer:

Option A

Step-by-step explanation:

This is a great question!

The first thing we want to do here is to graph the system of " inequalities " -

$\begin{bmatrix}2x+9y\le \:100\\ 9x+y\le \:54\end{bmatrix}$

As x is ≥ 0, respectively y ≥ 0, it makes things a lot simpler, as it kind of restricts us to quadrant 1, but not entirely.

First change each inequality to " slope - intercept " form, as such -

$2x + 10y \leq 100,\\10y \leq 100 - 2x,\\y \leq - 1 / 5x + 10\\----------\\9x + y \leq 54,\\y \leq - 9x + 54$

The graphed solutions would be in the attachment. I have colored the region with which the two intersect, and the fact that we are limited to quadrant 1. In this colored region there are 3 major points, ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ). We have to determine which of these are are maximum points, as the minimum are the same. Substitute these values ( ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ) ) into the equation f( x, y ) = 10x + 4y,

$f ( x, y ) = 10( 5 ) + 4( 9 ),\\f ( x, y ) = 86 -\\f ( x, y ) = 10( 0 ) + 4( 10 ),\\f ( x, y ) = 40 -\\f ( x, y ) = 10( 6 ) + 4( 0 ),\\f ( x, y ) = 60$

( 5, 9 ) resulted in the greatest amount, 86. That would make it our maximum point -

Solution = Option A

3 0

4 years ago