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olasank [31]
3 years ago
13

Write a recursive definition for the sequence 21,16,11,6

Mathematics
1 answer:
andrew11 [14]3 years ago
8 0
Hello :
6-11=11-16=16-21=-5
<span> the sequence is arithmetic r=-5

</span><span>a recursive definition is :  an+1 = an -5</span>
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Find an equation of the circle that satisfies the given conditions:
Anna71 [15]
<span>Given the two end of the diameter of the circle, we are able to compute the center of the circle as 0.5*[(-1,3)+(7,-7)]=(3,-2). The radius of the circle is 0.5*sqrt[(-1-7)^2+(3+7)^2]=sqrt(41). Therefore the equation of the circle is (x-3)^2+(y+2)^2=41.</span>
5 0
3 years ago
8 (1/2x -1/4) &gt; 12-2x
Deffense [45]

Answer:

x > 7/3

Step-by-step explanation:

Distribute by 8 first

4x - 2 > 12 - 2x

6x - 2 > 12

6x > 14

x > 14/6. or x > 7/3

3 0
3 years ago
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mario62 [17]

Answer:

  • y / cos x

Step-by-step explanation:

<u>Use trigonometry to help</u>

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7 0
3 years ago
Evaluate x(y-z)2 for =-1,y=5,and z=
Nata [24]

Answer:

The value of x(y-z)^2 is -16.

Step-by-step explanation:

We need to evaluate x(y-z)^2 for x = -1, y = 5 and let us assume that z = 1

It can be simply done by putting the values of x,y and z in the given expression.

x(y-z)^2=(-1)(5-1)^2\\\\=-1(4)^2

We know that the value of 4² = 16

So,

x(y-z)^2=16\times -1\\\\=-16

So, the value of x(y-z)^2 is -16.

6 0
4 years ago
Can I add <img src="https://tex.z-dn.net/?f=2%5Csqrt%7B3%7D%20%2B%205%5Csqrt%7B3%7D" id="TexFormula1" title="2\sqrt{3} + 5\sqrt{
Bond [772]

Yes, they can be added and simplified further. 2√3 + 5√3 = 7√3 .

Take √3 as common factor:

2√3 + 5√3

(2 + 5)√3

7√3

The two irrational numbers sums to form another irrational number 7√3 . In decimals that is 12.12435...

5 0
2 years ago
Read 2 more answers
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