Joshua uses a rule to write the following sequence of numbers 1/6,1/2,5/6,----------,11/2 What rule did Joshua use? What is the missing number in the sequence?

Answers

Answer 1

Answer: the numerator is added by 5 every other, and the denominator switches between 6 and 2

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Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x?

Answers

   5x = 6x² - 3
5x - 5x = 6x² - 5x - 3
       0 = 6x² - 5x - 3
   x = -(-5) ± √((-5)² - 4(6)(-3))
      2(6)
   x = 5 ± √(25 + 72)
      12
   x = 5 ± √(97)
   12
x = 5 + √(97) U x = 5 - √(97)
   12 12
$the\ values\ of\ x\ is\ (5 +/- √(97))/(12).$

The value of x are; x = 5 + √(97)/12 and x = 5 - √(97)/12.

What is a quadratic equation?

A quadratic equation is the second-order degree algebraic expression in a variable.

The standard form of this expression is ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

Given;

5x = 6x² - 3

Subtract 5x on both sides

5x - 5x = 6x² - 5x - 3

0 = 6x² - 5x - 3

x = -(-5) ± √((-5)² - 4(6)(-3)) / 2(6)

x = 5 ± √(25 + 72)/ 12

x = 5 ± √(97)/ 12

The value of x are;

x = 5 + √(97)/12

x = 5 - √(97)/12

Learn more about quadratic equations;

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In two or more complete sentences, explain the theorem used in solving for the range of possible lengths of the third side, AB of triangle ABC

Answers

Answer:

Step-by-step explanation:

In triangle if two sides are known and included angle is known we can use cosine formula as follows:

Say in a triangle, sides a,b are known and also included angle C

Then the third side

$c^(2) =a^(2)+b^(2) -2abCos C$

Since all values on right side are known, we can find the third side c easily.

Case II:

If alternately two sides and one angle not included is known. i.e we know a,b and either angle A or B.

then to find third side we use sine formula.

$(a)/(sinA)=(b)/(sinB) =(c)/(sinC)$

Using the above we can find the unknonwn side c easily.

You can use the Pythagorean Theorem to find the length of the third side AB (Identified as "x" in the figure attached in the problem), which says that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a² = b²+c²
As we can see the figure, the triangle does not have an angle of 90°, but it can be divided into two equal parts, leaving two triangles with a right angle. We already have the values of the hypotenuse and a leg in triangle "A" , so we can find the value of the other leg:
b = √(a²-c²) b = √(10²-4²) b = 9.16
With these values, we can find the hypotenuse in the triangle "B": x = √b²+c² x = √(9.16)²+(4)² x = 10

What is the perimeter of rectangle ABCD on the coordinate plane? coordinate grid

Answers

Drat it all I am so bad at reading. Sorry about that it is my fault :O

well the formula for a perimeter of a triangle is 2(lxw) so the length here is 10 i think (just count the cubes on the bottom) then with that number you multiply that by the width (side) then multiply by 2. i counted ten on the length and 5 on the width (im not sure if thats how any because its hard to cound on a tiny grid on a computer hehe), so i multiplied 10 x 5 and got 50, then multiply by 2, i got 100. im not sure, but ill double check :)

Find the volume of the square based pyramid.

Answers

A = 1/3 * s^2 * h where s is a side of the base and h is the height.

Finding the height of the pyramid:

$\left((6)/(2) \right )^(\!\!2)+h^(2)=8^(2)\n \n \n 3^(2)+h^(2)=8^(2)\n \n h^(2)=8^(2)-3^(2)\n \n h^(2)=64-9\n \n h^(2)=55\n \n h=√(55)\mathrm{~m}$

Finding the base area:

$S=6^(2)\n \n S=36\mathrm{~m^(2)}$

Formula for the volume of the pyramid:

$V=(S\cdot h)/(3)\n \n \n V=(36\cdot √(55))/(3)\n \n \n V=(\diagup\!\!\!\! 3\cdot 12\cdot √(55))/(\diagup\!\!\!\! 3)\n \n \n \boxed{\begin{array}{c} V=12√(55)\mathrm{~m^(3)} \end{array}}$

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n+1
4n
4n-4
O
4n
n+1
n
17+1
7+1
4
n+1