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2 years ago

Find the product of 3x and x^2-4x+10 subtract x^3+7x^2-2x

Mathematics

1 answer:

Virty [35]2 years ago

7 0

Answer:

below

Step-by-step explanation:

$2x^3 -5x^2 +28x$

I'm pretty sure this is the answer.

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1/4y + 1 1/2 + 2 − 1 3/4y − 12 = ?

pantera1 [17]

Brung like terms together:-
1/4 y - 1 3/4 y + 11/2 + 2 - 12
= - 1 1/2 y + 11/2 - 10
= - 1 1/2y - 9/2

8 0

3 years ago

Read 2 more answers

Consider the following polynomial.

vfiekz [6]

Answer:

The equivalent factored form of this equation is (x² + 49)(x - 6)

Step-by-step explanation:

x³ - 6x² + 49x - 294

First, group the first and second terms together and group the last two terms together.

(x³ - 6x²) + (49x - 294)

Find the greatest common factor of both parentheses and factor them.

x²(x - 6) + 49(x - 6)

Now, since the two terms in the parentheses are the same, then we have factored the equation correctly.

So, the factored form of the equation is (x² + 49)(x - 6)

3 0

2 years ago

Read 2 more answers

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$