Ask question

Ask question

All categories

2 years ago

11

Help me with this question please

1 answer:

Leokris [45]2 years ago

4 0

Answer:

14s^4-7s^2+15

Step-by-step explanation:

At the beginning the only bracket that is changing is the 3rd one because of the minus. (12s^4-6s^2+4s)+(6s^4-4s+27)-4s^4+s^2+12 then open the rest of the brackets. And start calculating.i will just to show you which ones u calculate together. (I'll separate the numbers depending on what they're squared to).

12s^4+6s^4-4s^4=14s^4

-6s^2-s^2=7s^2

27-12=15

All together is 14s^4-7s^2+15

You might be interested in

Two basketballs are thrown along different paths. Determine if the basketballs’ paths are parallel to each

Paul [167]

Answer:

Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.

Step-by-step explanation:

Remember that:

Two lines are parallel if their slopes are equivalent.
Two lines are perpendicular if their slopes are negative reciprocals of each other.
And two lines are neither if neither of the two cases above apply.

So, let's find the slope of each equation.

The first basketball is modeled by:

$\displaystyle 3x+4y=12$

We can convert this into slope-intercept form. Subtract 3<em>x</em> from both sides:

$4y=-3x+12$

And divide both sides by four:

$\displaystyle y=-\frac{3}{4}x+3$

So, the slope of the first basketball is -3/4.

The second basketball is modeled by:

$-6x-8y=24$

Again, let's convert this into slope-intercept form. Add 6<em>x</em> to both sides:

$-8y=6x+24$

And divide both sides by negative eight:

$\displaystyle y=-\frac{3}{4}x-3$

So, the slope of the second basketball is also -3/4.

Since the slopes of the two equations are equivalent, the basketballs' paths are parallel.

3 0

3 years ago

How to find the base and altitude of a rectangle if its area is 70 and its perimeter is 34?

kicyunya [14]

The area of a rectangle is base * altitude, and the perimeter is 2*base + 2*altitude,

base = 10cm and altitude = 7cm

6 0

3 years ago

State the conditions that the following graph meets. (Enter the equation or inequality that is shown by the graph.)

shutvik [7]

The graph shows a horizontal line intersecting (0, 2).

Your answer is: y = 2.

7 0

3 years ago

Read 2 more answers

△ABC is given with line m drawn through A parallel to BC¯¯¯¯¯¯¯¯. In the course of proving that the interior angle measures of △

aleksklad [387]

Answer:

I think a) and c is correct.

4 0

3 years ago

30+ Points!!!<br> 5. Solve the following inequalities.<br> a) 2 log3x – 2 logx3 -3 <0

Ilya [14]

Answer:

I answered your last question also

2 log3x – 2 logx3 -3 <0

$\mathrm{Subtract\:}2\log ^3\left(x\right)\mathrm{\:from\:both\:sides}$

$2\log ^3\left(x\right)-2logx^3-3-2\log ^3\left(x\right)$

$\mathrm{Simplify}$

$-2logx^3-3$

$\mathrm{Add\:}3\mathrm{\:to\:both\:sides}$

$-2logx^3-3+3$

$\mathrm{Simplify}$

$-2logx^3$

$Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)$

$\left(-2logx^3\right)\left(-1\right)>-2\log ^3\left(x\right)\left(-1\right)+3\left(-1\right)$

$\mathrm{Simplify}$

$2lx^3og>2\log ^3\left(x\right)-3$

$\mathrm{Divide\:both\:sides\:by\:}2lx^3o;\quad \:l>0$

$\frac{2lx^3og}{2lx^3o}>\frac{2\log ^3\left(x\right)}{2lx^3o}-\frac{3}{2lx^3o};\quad \:l>0\\$

$\mathrm{Simplify}$

$g>\frac{2\log ^3\left(x\right)-3}{2lx^3o};\quad \:l>0$

Step-by-step explanation:

6 0

3 years ago