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3 years ago

Dividing Fractions and finding the value of X

Advanced Placement (AP)

1 answer:

densk [106]3 years ago

7 0

Answer:

C.)0.5853267 ; D.)251.25169

Explanation:

3067/6730 + 9085/70097

0.4557206 + 0.1296061 = 0.5853267

205800/45097 + 98 -10 * 24 * 80050 - 100+1080 - 25x - 25x/254x

4.5634964 + 98 - 19212000 - 100 + 1080 - 25x - 25x/254x

C.)

345x + (-100)= 987x (- 60/216 +88)

345x - 100 = 987x(87.722222)

345x - 100 = 86581.833

345x = 86581.833 + 100

345x = 86681.833

x = 86681.833 / 345

x = 251.25169

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Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the

KIM [24]

Answer:

$T_{2020} = 15$

Explanation:

Given

$T_1 = 10$

$T_2 = 20$

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

This is proved as follows;

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

$T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1$

Assume n = 3

$T_{3} = \frac{T_1 + T_2}{2}$

Multiply both sides by 2

$2 * T_{3} = \frac{T_1 + T_2}{2} * 2$

$2T_{3} = T_1 + T_2$

Assume n = 4

$T_{4} = \frac{T_1 + T_2 + T_3}{3}$

$T_{4} = \frac{(T_1 + T_2) + T_3}{3}$

Substitute $2T_{3} = T_1 + T_2$

$T_{4} = \frac{2T_3 + T_3}{3}$

$T_{4} = \frac{3T_3}{3}$

$T_{4} = T_3$

Assume n = 5

$T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}$

$T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}$

Substitute $2T_{3} = T_1 + T_2$ and $T_{4} = T_3$

$T_{5} = \frac{2T_3 + T_3 +T_3}{4}$

$T_{5} = \frac{4T_3}{4}$

$T_{5} = \frac{(5-1)T_3}{5-1}$

Replace 5 with n

$T_{n} = \frac{(n-1)T_3}{n-1}$

(n-1) will definitely cancel out (n-1); So, we're left with

$T_{n} = T_3$

Hence,

$T_{2020} = T_3$

Calculating $T_3$

$T_{3} = \frac{10 + 20}{2}$

$T_{3} = \frac{30}{2}$

$T_{3} = 15$

Recall that $T_{2020} = T_3$

$T_{2020} = 15$

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3 years ago

In the period between 1820 and 1860, Southerners wanted slavery extended to the Western territories so that the South could

Scilla [17]

The answer is C. keep enough strength in the senate to protect southern interests

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In order to protect their own interest, they want slavery to extend to the north so they have enough strength in the senate to protect its legality

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melamori03 [73]

Answer: C. II, IV, I, III

Explanation:

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sdas [7]

Answer:

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Write the volume V of a cube as a function of the side length s

AveGali [126]

Answer:

v=s cube

The rate of change is found using the first derivative of this function. This is often called the gradient function, because it gives the gradient of a tangent line drawn at the specified point.

dv/ds(s cube) =3s square

We now plug in s=6

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Hope it's helpful to you

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