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miv72 [106K]
3 years ago
15

Subtract -2k3 + k2 - 9 from 5k3 - 3k + 7.

Mathematics
2 answers:
tankabanditka [31]3 years ago
8 0
5k3-3k+7-(-2k3+k2-9)
basically rewrite with simple algebra principles
15k-3k+7-(-6k+2k-9)
split up the parenthesis
15k-3k+7+6k-2k+9
sort them
15k-3k+6k-3k+7+9
short down by adding up the similar factors
answer: 18k+18
factorise
18(k+1)
both forms are right
Sever21 [200]3 years ago
7 0

Answer:  The required value is 7k^3-k^2-3k+16.

Step-by-step explanation:  We are given to subtract the polynomial -2k^3+k^2-9 from 5k^3-3k+7.

That is, we need to find the value of S, where

S=(5k^3-3k+7)-(-2k^3+k^2-9)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

To perform the given subtraction, we need to combine the like terms as follows :

S\\\\=(5k^3-3k+7)-(-2k^3+k^2-9)\\\\=5k^3-3k+7+2k^3-k^2+9\\\\=(5+2)k^3-k^2-3k+(9+7)\\\\=7k^3-k^2-3k+16.

Thus, the required value is 7k^3-k^2-3k+16.

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Please help.. with 6 & 15 ​
Black_prince [1.1K]

Answer:

6. C: {x^2 +(y-1)^2 =2; x+y = 3}

15. C: The line does not intersect the circle.

Step-by-step explanation:

The formula for the distance (d) from a point (x, y) to a line ax+by=c is ...

d = |ax+by-c|/√(a^2+b^2)

The formula for a circle centered at (h, k) with radius r is ...

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6. Comparing the circle equation to the generic equation, we find (h, k) = (0, 1) and r = √2. Then we want to find the line that is distance √2 from the center of the circle. Our line equation is x+y=c for some value of c that we want to find.

d = √2 = |0 +1 -c|/√(1^2+1^2)

2 = |1-c|

±2 = 1-c

c = 1±2 = -1 or 3

The line that is tangent to the circle is the one of choice C: x+y = 3

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The attached graph shows the lines for all 4 answer choices. The point of tangency is (1, 2), so x+y=1+2=3.

___

15. The circle is centered at (4, 1) and has radius 3. The distance from the circle center to the line is ...

d = |2(4) -(1)|/√(2^2+(-1)^2) = 7/√5 ≈ 3.13

The distance from the circle center to the line is more than the radius of the circle, so there can be no points of intersection.

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Alternate solution

You can substitute for y using the equation of the line. Then the circle equation becomes ...

(x -4)^2 + (2x -1)^2 = 9

x^2 -8x +16 +4x^2 -4x +1 = 9

5x^2 -12x +8 = 0

The discriminant of this quadratic is ...

b^2 -4ac = (-12)^2 -4(5)(8) = 144-160 = -16

Since this value is negative, there can be no real solutions, meaning the line does not intersect the circle.

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3 years ago
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frez [133]

Answer:

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3 0
3 years ago
2x (y^2+3)-5 (3+y^2)
Irina18 [472]

Use distributive property a(b + c) = ab + ac.

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