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2 years ago

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Sarah would like to make a 6 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of a mixture that is 80%

peanuts and 20% almonds and several pounds of a mixture that is 50% peanuts and 50% almonds.
a. What is the system that models this situation?
b. What is the solution to the system? How many pounds of the 80/20 mixture? How many pounds of the 50/50 mixture?

1 answer:

ivolga24 [154]2 years ago

7 0

Answer:

<em>a. The system of equations that models the situation is....</em>

<em> $0.80x+0.50y=3.60\\\\ 0.20x+0.50y=2.40$ </em>

<em>b. The solution to the system: x = 2 and y = 4 </em>

<em>The amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.</em>

Step-by-step explanation:

Suppose, the amount of 80/20 mixture is $x$ pounds and the amount of 50/50 mixture is $y$ pounds.

So, the amount of peanuts in 80/20 mixture $= 0.80x$ pound and the amount of almonds in 80/20 mixture $=0.20x$ pound.

And the amount of peanuts in 50/50 mixture $=0.50y$ pound and the amount of almonds in 50/50 mixture $=0.50y$ pound.

Now, Sarah would like to make a 6 pounds nut mixture that is 60% peanuts and 40% almonds.

So, the amount of peanuts in that mixture $=(6\times 0.60)=3.60$ pounds

and the amount of almonds in that mixture $=(6 \times 0.40)= 2.40$ pounds.

So, the system of equations will be.........

$0.80x+0.50y=3.60 ...................(1)\\\\ 0.20x+0.50y=2.40...................(2)$

Subtracting equation (2) from equation (1), we will get.....

$(0.80x+0.50y)-(0.20x+0.50y)=3.60-2.40\\ \\ 0.60x=1.20\\ \\ x= \frac{1.20}{0.60}=2$

Now, plugging this $x=2$ into equation (1), we will get......

$0.80(2)+0.50y=3.60\\ \\ 1.60+0.50y=3.60\\ \\ 0.50y=3.60-1.60=2\\ \\ y=\frac{2}{0.50}=4$

So, the amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.

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Answer:

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Step-by-step explanation:

1st digit can have the values 1-9 (9 distinct values)

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2 years ago

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1 year ago

In one year the perseid metor shower had a metor appear every 1/5 5 minutes on average That same year the Leonid metor shower ha

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In one year, the Perseid meteor shower had a meteor appear every 1 1/5 minutes on average. That same year, the Leonid meteor shower had a meteor appear every 4 2/3 minutes on average. How many more meteors fell during the Perseid meteor shower?

Answer:

325,452 meteors

Step-by-step explanation:

We all know that :

We have 365 days in a year and in each day there are 24 hours, Likewise an hour has 60 minutes

SO, the total number of minutes in a year is :

365 × 24 × 60 = 525600 minutes

For Perseid meteor:

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So one meteor fall at 1.2 minutes interval Thus, in a year, we will have :

1 meteor - 1.2 minutes.

x meteor - 525600 minutes.

1.2x = 525600

$x = \frac{525600}{1.2}$

x = 438000

∴ 438,000 meteors fell during the Perseid shower.

For Leonid meteor shower:

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So one meteor fall at 4.67 minutes interval Thus, in a year, we will have :

1 meteor - 4.67 minutes.

x meteors - 525600 minutes.

$4.67x = 525600$

$x = \frac{525600}{4.67}$

x = 112548

112,548 meteors fell during the Leonid Shower.

Finally , the numbers of more meteors that fell during the Perseid meteor shower is calculated by the difference in the number of Perseid meteor shower and Lenoid meteor shower. i.e

438,000 - 112,548 = 325,452

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2 years ago

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2 years ago

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 $r$ :

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

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

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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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2 years ago