For which pairs of functions is (f times g)(x)=x? Please help!!!!

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 20 N acts on

jeka94

The acceleration of the object will be 10 m/s²

Step-by-step explanation:

Direct variation is a relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other

If y varies directly with x, then y ∝ x
y = k x, where k is the constant of variation

For a moving object, the force acting on the object varies directly with the object's acceleration.

Assume that the force is F and the acceleration is a

∵ F ∝ a

∴ F = k a

∵ F = 20 newtons

∵ a = 4 m/s²

- Substitute these values in the equation above to find k

∵ 20 = k (4)

∴ 20 = 4 k

- Divide both sides by 4

∴ k = 5

- Substitute the value of k in the equation

∴ F = 5 a ⇒ equation of variation

∵ F = 50 Newtons

∵ F = 5 a

∴ 50 = 5 a

- Divide both sides by 5

∴ 10 = a

∴ a = 10 m/s²

The acceleration of the object will be 10 m/s²

Learn more:

You can learn more about variation in brainly.com/question/10708697

#LearnwithBrainly

7 0

2 years ago

Which points are the verticles of the ellipse? Equations of ellipse

Mashutka [201]

Answer:

This question is solved in detail below. Please refer to the attachment for better understanding of an Ellipse.

Step-by-step explanation:

In this question, there is a spelling mistake. This is vertices not verticles.

So, I have attached a diagram of an ellipse in which it is clearly mentioned where are the vertices of an ellipse.

Vertices of an Ellipse: There are two axes in any ellipse, one is called major axis and other is called minor axis. Where, minor is the shorter axis and major axis is the longer one. The places or points where major axis and minor axis ends are called the vertices of an ellipse. Please refer to the attachment for further clarification.

Equations of an ellipse in its standard form:

$\frac{x^{2} }{a^{2} } + \frac{y^{2} }{b^{2} } = 1$

This is the case when major axis the longer one is on the x-axis centered at an origin.

$\frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} } = 1$

This is the case when major axis the longer one is on the y-axis centered at an origin.

where major axis length = 2a

and minor axis length = 2b

3 0

2 years ago

Solve for x.

Masteriza [31]

Answer:

x = 2.5 or -0.8

Step-by-step explanation:

Here, we are to use completing the square method to solve for the values of x

Firstly, we divide through by 4

= x^2 -7/4x -2 = 0

Now, we move the two term to the right hand side

x^2 -7/4x = 2

Now, we divide the coefficient of x by 2 and square it;

That would be;

(-7/4 * 1/2)^2 = 49/64

we now add this value to both sides of the equation

x^2 -7/4x + 49/64 = 2 + 49/64

The right hand side can be rewritten as;

(x-7/8)^2 = 177/64

taking the square root of both sides

x-7/8 = √(177/64)

x = 7/8 ± √(177/64)

x = 7/8 + √(177/74) or 7/8 - √(177/64)

= x = 2.53802 or -0.788017

Which to the nearest tenth is

x = 2.5 or -0.8

7 0

3 years ago

Segments

Kazeer [188]

Given

AB and CD intersect

AC, CB, BD and AD are congruent.

Prove that AB is the bisector of ∠CAD and ray CD is the bisector of ∠ACB.

and AB and CD are perpendicular.

To proof

Bisector

<em>A bisector is that which cut an angle in two equal parts.</em>

In ΔACB and ΔADB

AD = AC ( Given )

AB = AB ( common )

BC = DB ( Given )

by SSS congurence property

we have

ΔACB ≅ΔADB

∠CAB =∠ DAB

∠CBA = ∠DBA

( By corresponding sides of the congurent triangle )

Thus AB is the bisector of the ∠CAD.

InΔ DAC and ΔDBC

AD = DB (Given)

AC = CB ( Given )

CD = CD (common)

By SSS congurence property

ΔDAC≅ Δ DBC

∠ ACD =∠ BCD

∠ADC =∠BDC

( By corresponding sides of the congurent triangle )

Therefore CD is the bisector of the CAD.

In ΔBOC andΔ BOD

BO = BO ( Common )

∠BCO = ∠BDO

( As prove above ΔACB ≅ΔADB

Thus ∠ACB = ∠ADB by corresponding sides of the congurent triangle , CD is a bisector

∠BCO = ∠BDO )

CB = DB ( given )

by SAS congurence property

ΔBOC ≅ ΔBOD

∠BOC =∠ BOD

∠BOC +∠ BOD = 180 °( Linear pair )

2∠ BOC = 180°

∠BOC = 90°

∠BOC =∠ BOD = 90°

also

In ΔCOA and ΔAOD

AO = AO ( Common )

∠ACO =∠ ADO

( As prove above ΔACB ≅ΔADB Thus ACB = ADB by corresponding sides of congurent triangle ,CD is a bisector

thus ∠ACO = ∠ADO )

AC =AD ( given )

by SAS congurence property

Δ COA ≅ ΔAOD

∠AOC = ∠AOD

( By corresponding angle of corresponding sides )

∠AOC + ∠AOD = 180°

2∠ AOC = 180° ( Linear pair )

∠AOC = 90°

∠AOC = ∠AOD = 90 °

Thus AB and CD are perpendicular.

Hence proved

6 0

3 years ago

Find the fourth-degree polynomial function with zeros 4, -4, 4i , and -4i . Write the function in factored form.

Iteru [2.4K]

Given:

A fourth-degree polynomial function has zeros 4, -4, 4i , and -4i .

To find:

The fourth-degree polynomial function in factored form.

Solution:

The factor for of nth degree polynomial is:

$P(x)=(x-a_1)(x-a_2)...(x-a_n)$

Where, $a_1,a_2,...,a_n$ are n zeros of the polynomial.

It is given that a fourth-degree polynomial function has zeros 4, -4, 4i , and -4i. So, the factor form of given polynomial is:

$P(x)=(x-4)(x-(-4))(x-4i)(x-(-4i))$

$P(x)=(x-4)(x+4)(x-4i)(x+4i)$

$P(x)=(x-4)(x+4)(x^2-(4i)^2)$ $[\because a^2-b^2=(a-b)(a+b)]$

On further simplification, we get

$P(x)=(x-4)(x+4)(x^2-4^2i^2)$

$P(x)=(x-4)(x+4)(x^2+16)$ $[\because i^2=-1]$

Therefore, the required fourth degree polynomial is $P(x)=(x-4)(x+4)(x^2+16)$ .

6 0

2 years ago