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Kitty [74]
3 years ago
12

Find the product. Multiplication of the binomial.

Mathematics
1 answer:
AnnZ [28]3 years ago
6 0

We have, (2r+3)(3r^2+4r+5) or more generally,

(a+b)(c+d+e)=ac+ad+ae+bc+bd+be

So using that, we find,

(2r+3)(3r^2+4r+5)=6r^3+8r^2+10r+9r^2+12r+15, then collect terms with same exponent and get \boxed{6r^3+17r^2+22r+15}.

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X=41 Just S=ubract The 4 from 37 Then Subract X From X And Divide The Whole Thing By -1
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2 years ago
What is the equation of a line passing through (-3, 7) and having a slope of ?
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The slope cannot be found when only one co-ordinate is given.

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This answer is ambiguous
5 0
3 years ago
Calculate the radius of a circle that has a diameter of 190
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Answer:

30.23944

Step-by-step explanation:

5 0
3 years ago
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Write an equation of the circle that has a diameter with endpoints (12, 3) and<br> (-18,3).
Mademuasel [1]

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

<h3>Coordinates of center of circle</h3>

Since the endpoints of the diameter are (x₁, y₁) = (12,3) and (x₂, y₂) = (-18,3), the coordinates of the center of the circle are the midpoints of the diameter. So, the midpoints are

  • x = (x₁ + x₂)/2 = (12 + (-18))/2 = (12 - 18)/2 = -6/2 = -3 and
  • y = (y₁ + y₂)/2 = (3 + 3)/2 = 6/2 = 3.

So, the coordinates of the center of the circle are (-3, 3)

<h3>The radius of the circle</h3>

The radius of the circle r = √[(x₁ - h)² + (y₁ - k)²] where

  • (x₁, y₁) = coordinates of end of diameter = (12, 3) and
  • (h, k) = coordinates of center of circle = (-3, 3)

So, substituting the values of the variables into r, we have

r = √[(x₁ - h)² + (y₁ - k)²]

r = √[(12 - (-3))² + (3 - 3)²]

r = √[(12 + 3)² + 0²]

r = √[15² + 0²]

r = √15²

r = 15

<h3>The equation of the circle</h3>

The equation of a circle with center (h,k) is given by

(x - h)² + (y - k)² = r² where r = radius of circle.

Substituting the values of the variables into the equation, we have

(x - h)² + (y - k)² = r²

(x - (-3))² + (y - 3)² = 15²

(x + 3)² + (y - 3)² = 15²

x² + 6x + 9 + y² - 6y + 9 = 225

x² + 6x + y² - 6y + 9 + 9 - 225 = 0

x² + y² + 6x - 6y - 207 = 0

The equation of the circle that has a diameter with endpoints (12, 3) and(-18,3) is x² + y² + 6x - 6y - 207 = 0

Learn more about equation of a circle here:

brainly.com/question/18435467

5 0
2 years ago
Evaluate pq÷4; use p=14, and q=14​
Zolol [24]

Answer:

49

Step-by-step explanation:

14x14 = 196

196/4 = 49

4 0
3 years ago
Read 2 more answers
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