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3 years ago

Find the sum of the first 12 terms of the sequence. Show all work for full credit.

Mathematics

1 answer:

xeze [42]3 years ago

6 0

Answer:

The sum of 12 terms of AP = -318

Step-by-step explanation:

Points to remember

Sum of n terms of an AP

Sₙ = n/2[ 2a + (n - 1)d]

Where n - number of terms

a - first term

d - common difference

To find the sum of 12 terms

The given AP is 1, -4, -9, -14, . . .

a = 1, d= -5 and n = 12

Sₙ = n/2[ 2a + (n - 1)d]

= 12/2[2*1 + (12 - 1)*(-5)]

= 6[2 + (11*(-5))]

= 6[2 - 55]

= 6 * (-53)

= -318

The sum of 12 terms of AP = -318

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soldi70 [24.7K]

Answer:

d. 120 words

Step-by-step explanation:

This question can be solved by a simple rule of three.

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1 minute - 40 words

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So the correct answer is:

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5 0

3 years ago

1. Find the volume for 5 different spheres by randomly choosing different radii.

Sophie [7]

The volume of the sphere is given by :

V= $\frac{4}{3}\pi r^{3}$

And volume of cylinder is given as :

V= $\pi r^{2}h$

So, as given to select the radius randomly and the height of the cylinder will be the diameter of the sphere.

The radius are = 3, 5, 8, 10, 12

And height will be = 6, 10, 16, 20, 24

1. Volume of sphere with radius 3 = $\frac{4}{3}*3.14*3*3*3= 113.04$ units

Volume of cylinder with radius 3 and height 6 = $3.14*3*3*6=169.56$ units

2. volume of sphere with radius 5 = 523.60 units

volume of cylinder with height 10 = 785.40 units

3. volume of sphere with radius 8 = 2144.66 units

volume of cylinder with height 16 = 3216.99 units

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3 0

3 years ago

The sum of four siblings ages is 71. The second child’s age is twice the youngest, the third child’s age is twelve more than the

user100 [1]

Answer: 9, 18, 21, and 23 respectively

Step-by-step explanation:

To solve any problem like this, you want to assign a variable to each child because you dont know their age. So lets write an equation and assign variables to these children.

r = first child (youngest)

x = second child

y = third child

z = fourth child (oldest)

We know that r + x + y + z = 71

We also are given info on the comparisons of their ages.

r = r

x = 2r (2 times the youngest)

y = 12 + r (12 more than the youngest)

z = 3r - 4 (4 less than 3 times the youngest)

That was the easy part tbh, now we have to solve for 'r'. We can see that all the ages are a function of the same variable (r), and we know the sum of them is equal to 71. So lets solve for r and then plug that r value into each equation we just derived.

Child #1 (youngest)

$r+2r+(12+r)+(3r-4)=71\\3r+12+r+3r-4=71\\7r+12-4=71\\7r+8=71\\7r=63\\r=9yrs$