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Novosadov [1.4K]
3 years ago
5

Order the polynomial in descending powers of r . Do not repeat terms.

Mathematics
2 answers:
VikaD [51]3 years ago
6 0
Placing these in descending order we use the exponents to do this...

-4r³s - 3r² + 2rs - 5
poizon [28]3 years ago
5 0

Answer:

The required order of polynomial is -4r^3s-3r^2+2rs-5.

Step-by-step explanation:

The given polynomial is

2rs-5-3r^2-4r^3s

We have to arrange the polynomial in descending powers of r.

In term 2rs, the power of r is 1.

In term -5, the power of r is 0.

In term -3r², the power of r is 2.

In term -4r³s, the power of r is 3.

On arranging the polynomial in descending powers of r, we get

-4r^3s-3r^2+2rs-5

Therefore the required order of polynomial is -4r^3s-3r^2+2rs-5.

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Write -1/3c^2(1.2d^2-6c) as a polynomial
denis23 [38]

Answer:

-0.4c^2d^2 + 2c^3

Step-by-step explanation:

Multiply -1/3c^2 with both terms inside the parentheses to get the polynomial.

<em>mark as brainliest if it helped</em>

5 0
2 years ago
What’s the area???????
Damm [24]

Answer:

48 cm

Step-by-step explanation:

hope this helps!

6 0
3 years ago
What is the answer???<br> How can we solve it?
GenaCL600 [577]
Okay, so it's pretty easy.
Do 11*11=121

times that by 1/2

so 60.5
v=60.5(6)
now times that by 6
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4 0
3 years ago
Read 2 more answers
Which is not an equation of the line going through (6,7) and (2,-1) A. y-7=2(x-6) B. y=2x-5 C. y-1=2(x+2) D. y+1=2(x-2)
nikdorinn [45]

Answer:

C. y-1=2(x+2)

Step-by-step explanation:

A. y-7=2(x-6)-point slope form: y-y1=2(x-x1)-coordinate point: (x1,y1)=(6,7)

Now let's check that the other point works-just sub in the coordinates for the x and y variables.

y-7=2(x-6)\\-1-7=2(2-6)\\-8=2(-4)\\-8=-8

So answer choice A. goes through both points, so no.

B. y=2x-5-slope-intercept form, so just sub in both sets of coordinates separately to see if they each work.

y=2x-5\\7=2(6)-5\\7=12-5\\7=7\\equal

y=2x-5\\-1=2(2)-5\\-1=4-5\\-1=-1\\equal

Both work, so no.

C. y-1=2(x+2)-point slope form that gives us first coordinate set--(-2,1)

Except that the point is actually (2,-1) not (-2,1).  That means that answer choice C is not an equation of the line thaat goes through (6,7) and (2,-1), so yes.

D. Anyway, if we check D--y+1=2(x-2)--coordinate point given: (2,-1), which is correct, so no.

4 0
3 years ago
HELP ME PLEASE!!!!!!!!!
shepuryov [24]
I think the answer might be T(x) = (t+8)(t-3).
This will allow you to easily figure out what values of t will make T(x)=0. If t = -8, or t = 3, you will get zero for an answer
3 0
3 years ago
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