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Alexandra [31]
2 years ago
7

Find the value of k so that the remainder is 1 when x^3+5x^2+kx-8 is divided by (x-3) K=

Mathematics
1 answer:
Brut [27]2 years ago
5 0

By the remainder theorem, the remainder upon dividing x^3+5x^2+kx-8 by x-3 is given by the value of this expression when <em>x</em> = 3 :

3^3 + 5\cdot3^2 + 3k - 8 = 1

Solve for <em>k</em> :

27 + 45 + 3k - 8 = 1 \\\\ 3k = -63 \\\\ \boxed{k = -21}

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4 0
3 years ago
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8 0
3 years ago
If line AD is a tangent to circle B at point C, and m ABC = 55°, what is the measure of BAD?
nikdorinn [45]

Answer:

145°

Step-by-step explanation:

There are a couple of ways you can get there:

1. ∠ACB is a right angle, 90°. Hence, ∠BAC is the complement of ∠ABC, so is ...

... ∠BAC = 90° -∠ABC = 90° -55° = 35°

Then, ∠BAC and ∠BAD are a linear pair, so total 180°. That makes ∠BAD the supplement of ∠BAC, so ...

... ∠BAD = 180° -35° = 145°

2. ∠BAD is the exterior angle at A for the triangle ABC. It will have a measure that is the sum of the opposite interior angles: given ∠ABC = 55° and right angle ACB = 90°.

... ∠BAD = 55° +90° = 145°

8 0
3 years ago
Help me plz!!!!!!!!!!!!!
4vir4ik [10]
22 and 27 is the answer. what else do you need??
8 0
4 years ago
How can you use the multiplication property of equality to rewrite the equation 0.6x + 4.8 = 7.2
Kobotan [32]
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0.6x=7.2-4.8 

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7 0
3 years ago
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