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3 years ago

9

Joe and Ben are saving money for a new x-box game. Joe has saved $15.00 in 2 weeks and $37.50 in 5 weeks. Ben's savings are repr

esented by the equation
y=8.25x
y=8.25x
.
Find the difference in their savings each week?

1 answer:

ivanzaharov [21]3 years ago

5 0

Answer:

37.50/5 = $7.50/week

so the difference is 8.25-7.50 = $0.75/week

Step-by-step explanation:

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What is the approximate 47? Give your answer and explain why it is correct. HELP

a_sh-v [17]

Answer:

7

Step-by-step explanation:

7 is approximate to $\sqrt{47}$ because 7 is $\sqrt{49}$ , and 49 is close to 47

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Read 2 more answers

Factor 140c + 28 - 14a to identify the equivalent expressions.

Feliz [49]

Answer:

The answer to your question are letters A, C and D.

Step-by-step explanation:

To factor this polynomial we need to look for the common factor of the three terms.

To find it, get the greatest common factor

140 28 14 2

70 14 7 2

35 7 7 5

7 7 7 7

1 1 1

Greatest common factor = 2 x 7 = 14

Factor the polynomial

140c + 28 - 14a = 14 ( 10c + 2 - a)

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David and Mark are marking exam papers. Each set takes David 24 minutes and Mark 1 hour. Express the times David and Mark take a

attashe74 [19]

Answer:

2:5

Step-by-step explanation:

david=24mins

mark = 1hour

change 1 hour to mins

DAVID=24

MARK=60

24:60

THEN SIMPLIFY

24/60

<em>=</em><em>2</em><em>/</em><em>5</em>

<em><u>2</u></em><em><u>:</u></em><em><u>5</u></em>

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In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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