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Bingel [31]
3 years ago
15

(3x+2) (-6x-3) quadratic and linear and constant term

Mathematics
1 answer:
gavmur [86]3 years ago
8 0
9x + 5 is the answer
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Simplify a2b3×a5b2​
Kryger [21]

Answer:

\huge \orange { {a}^{7}  {b}^{5} }

Step-by-step explanation:

\huge {a}^{2}  {b}^{3}  \times  {a}^{5}  {b}^{2}  \\  \\\huge =  {a}^{2}  \times  {a}^{5}   \times {b}^{3}  \times  {b}^{2}  \\ \\\huge =  {a}^{2 + 5}  \times  {b}^{3 + 2}  \\ \\  \huge \red{=  {a}^{7}  {b}^{5} }

6 0
3 years ago
An email spam reaches three readers. Each of them forwarded the same email to three different readers. Each of these new readers
SOVA2 [1]

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Step-by-step explanation:

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3 years ago
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Find the angle of the sector indicated. Find the correct answer to the problem.
svlad2 [7]

The correct answer is (C)!

Since the area of the circle is 400π, 150π/400π = 135/360.

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3 years ago
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What is 10⁰ + 10¹ + 10² equal to
Setler [38]
10^0 = 1

10^1 = 10

10^2 = 100

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6 0
3 years ago
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The universal set in this diagram is the set of integers from 1 to 15. place the integers in the correct place in the venn diagr
neonofarm [45]

Answer:


Step-by-step explanation:

Given :  universal set in this diagram is the set of integers from 1 to 15.

Solution :

The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15

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The intersection of odd integer,multiples of 3 is 9

The remaining multiples of 3 are 6,12

The remaining odd integers are 7,11,13

Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .

Refer the attached figure for the answer.

3 0
3 years ago
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