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3 years ago

Check all the answers that are more precise than a weight of 200 grams.

Mathematics

1 answer:

Damm [24]3 years ago

8 0

Answer:

amigo mío la respuesta es la letra B

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Solve the following equation, answer as a reduced, mixed number. Then place the correct number in the box provided. 4(7 - 3x) =

Luda [366]

Answer: x = 0

Step-by-step explanation:

4(7 - 3x) = 7(4 - 2x)

28 - 12x = 28 - 14x distributed

28 + 2x = 28 added 14x to both sides

2x = 0 subtracted 28 from both sides

x = 0 divided both sides by 2

3 0

3 years ago

Solve each of the following equations by the trial and error method : A) X- 7= -3 help some one now

Vaselesa [24]

Answer:

X - 7 = - 3

X = - 3 + 7

X = 4

the 7 turns positive because it was moved to the other side of the symbol "=" so we reverse the sign

- becomes +

and

+ becomes -

8 0

3 years ago

Suppose that the credit remaining on a phone card (in dollars) is a linear function of the total calling time (in minutes). When

Virty [35]

Answer:

Credit remaining after 21 minutes = $30.4

Step-by-step explanation:

Credit remaining on a phone card is a linear function of the total calling time.

When graphed, let the linear function representing the line is,

y = mx + b

Where 'm' = slope of the line

b = y-intercept

From the graph,

Slope of the line = -0.12

y = -0.12x + b

If this line passes through a point (33, 28.96),

28.96 = -0.12(33) + b

b = 28.96 + 3.96

b = 32.92

Therefore, the linear function is,

f(x) = -0.12x + 32.92

where x = calling time

Credit left in the card after 21 minutes,

f(21) = -0.12(21) + 32.92

= -2.52 + 32.92

= $30.4

6 0

3 years ago

A little help here? Whats the value of X?

zubka84 [21]

You’d add 65+79 to get 144 then subtract that from 180 and get 36 so X would be 36 :) hope this helped!

4 0

3 years ago

Read 2 more answers

Solve the following recurrence relation: <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

-Dominant- [34]

By iteratively substituting, we have

$a_n = a_{n-1} + n$

$a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)$

$a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)$

and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

It follows that

$a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2$

$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

4 0

2 years ago