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3 years ago

F(x)=(x^3+4x^2+7x-9) d(x)=(x+3) divide using long division

Mathematics

1 answer:

Alchen [17]3 years ago

7 0

The result is x^2 +x +4 -21/(x+3).

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Polynomial long division tends to be easier than numerical long division because the quotient term never needs adjustment. It is always the ratio of the highest degree terms of the dividend and divisor.

See the attachment for the intermediate steps.

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Solve for x. Assume that lines which appear tangent are tangent.

REY [17]

Remark
Secants going through a circle form a ratio that's the same for all secants going through the circle and beginning at 1 point. See the diagram below for the ratio. Follow the Line designations carefully. This comes up many times.

General Ratio is (segment from the external point to the the first intersect point of the circumference divided by the distance from the external point to the second intercept point on the circumference).

General Ratio
Put in terms of the diagram the general ratio is (OA / OB)

Equation
OA/OB = OC/OD

Substitute and Solve
9/(9 + 11) = 10/(10 + x)
9/20 = 10/(10 + x) Notice that you can't combine the right side. Cross Multiply

9*(10 + x) = 10 * 20 Remove the brackets on the left. Combine the right factors.
90 + 9x = 200 Subtract 90 from both sides.
9x = 200 - 90
9x = 110 Divide by 9
110 / 9 = 12.22

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3 years ago

What is the common value for all of these expressions?<br> 4.5 / 0.09 45 / 0.9 450 / 9 4,500 / 90

Rasek [7]

Four, five and nine……

3 0

2 years ago

What is the value of x in the equation 4x + 1.3 = -7.9

forsale [732]

Answer:

x = - 2.3

Step-by-step explanation:

4x + 1.3 = -7.9
4x = -7.9 - 1.3
4x = -9.2
x = -9.2/4
x = - 2.3

4 0

3 years ago

Read 2 more answers

What is it called when an example is showing a statement is not true

lorasvet [3.4K]

If you show that the example is showing a false statement then you are disproving the example/problem.

5 0

3 years ago

A uniform 41.0 kg scaffold of length 6.6 m is supported by two light cables, as shown below. A 74.0 kg painter stands 1.0 m from

Nimfa-mama [501]

Step-by-step explanation:

Let

$m_p$ = mass of the painter

$m_s$ = mass of the scaffold

$m_e$ = mass of the equipment

$T$ = tension in the cables

In order for this scaffold to remain in equilibrium, the net force and torque on it must be zero. The net force acting on the scaffold can be written as

$3T = (m_p + m_s + m_e)g\:\:\:\:\:\:\:(1)$

Set this aside and let's look at the net torque on the scaffold. Assume the counterclockwise direction to be the positive direction for the rotation. The pivot point is chosen so that one of the unknown quantities is eliminated. Let's choose our pivot point to be the location of $m_e$ . The net torque on the scaffold is then

$T(1.4\:\text{m}) + m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m}) - 2T(5.2\:\text{m}) = 0$

Solving for T,

$9T = m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})$

$T = \frac{1}{9}[m_sg(1.9\:\text{m}) + m_pg(4.2\:\text{m})]$

$\:\:\:\:= 423.3\:\text{N}$

To solve for the the mass of the equipment $m_e$ , use the value for T into Eqn(1):

$m_e = \dfrac{3T - (m_p + m_s)g}{g} = 14.6\:\text{kg}$

6 0

3 years ago