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schepotkina [342]

3 years ago

15

Scott is reading two books at the same time to prepare for a report he is writing. So far, he has read 407 of the total number o

f 557 pages, which is Three-fifths of the shorter book and StartFraction 5 Over 6 EndFraction of the longer book. Which system of equations can be used to determine the total number of pages in the shorter book, x, and the total number of pages in the longer book, y?

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

The system of equations are

x+y=557

$\frac35x+\frac56 y=407$

The total pages in shorter book is 245

The total pages in longer book is 312

Step-by-step explanation:

Given that,

Scott has read 407 of the total number of 557 pages, which $\frac35$ of the shorter book and $\frac56$ of longer book.

Total number of pages of shorter book be x and longer book be y.

Then,

$\frac35$ page of the shorter book $=\frac35 x$

$\frac56$ pages of the longer book = $\frac56y$

So, he has read $=\frac35x+\frac 56y$

Total number of pages of both book is = x+y

According to the problem,

x+y=557.........(1)

$\frac35x+\frac56 y=407$ .......(2)

We can write equation (2) as

$\frac35x+\frac56 y=407$

$\Rightarrow \frac{18x+25y}{30}=407$

$\Rightarrow {18x+25y}=407\times 30$

$\Rightarrow {18x+25y}=12,210$ ......(3)

Now 18 times of equation (1) subtract from equation (3)

18x+25y=12210

18x+18y=10026

- - -

_______________

25y-18y=12,210-10,026

⇒7y=2,184

$\Rightarrow y=\frac{2184}{7}$

⇒y= 312

Plug y=312 in equation (1)

x+312=557

⇒x=557-312

⇒x=245

The total pages in shorter book is 245

The total pages in longer book is 312

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Here, thankyou!!!!!!!

FromTheMoon [43]

Answer:

Perimeter of i - 22CM

Area of i - 13CM^2

Perimeter of ii -38CM

Area of ii -66CM^2

Perimeter of iii -30CM

Area of iii- 42CM^2

Perimeter of iv - 50CM

Area of iv- 126CM^2

Step-by-step explanation:

SHAPE I:

PERIMETER = S + S + S + S + S +S

= 7 + 1 + 5 +3 +4 +2

= 22CM

AREA = Part a - 4 * 2 = 8cm^2 part B - 5 *1 = 5cm^2

Total = 8 + 5 = 13cm^2

SHAPE II:

PERIMETER = S + S + S + S + S +S

= 4 + 4 +5 + 6 + 9 + 10

= 38 CM

AREA = Part a - 5 * 10 = 50 cm^2 part B - 4 *4 =16 cm^2

Total = 50+16 =66 cm^2

SHAPE III:

PERIMETER = S + S + S + S + S +S

= 9 + 2 + 3 + 4 + 6 + 6

= 30CM

AREA = Part a - 6 * 6 = 42cm^2 part B - 3 * 2= 6cm^2

Total = 36 + 6 =42 cm^2

SHAPE IV:

PERIMETER = S + S + S + S + S +S

= 9 + 10 + 4 + 6 + 6 + 15

= 50 CM

AREA = Part a -15 * 6 = 90 cm^2 part B - 9 *4 = 36cm^2

Total = 90 + 36 = 126cm^2

HOPE THIS HELPED

6 0

3 years ago

300 pounds of oranges were purchased at $0.24 per pound. The desired markup is 50% based on selling price, but 21% spoilage is e

andrezito [222]

1. The problem indicates that 300 pounds of oranges were purchased at $0.24 per pound, so the cost is:

300x$0.24=$72

2. A percent of 21% spoilage is expected, which means that 89% of 300 pounds of oranges left, is:

300x0.89=267 pounds of oranges

3. We want to calculate the selling price per pound of 267 pounds of oranges, so let's call this value "x":

Markup %=(Markup/cost)x100

Markup=267x-72

4. The desired markup, based on selling price, is 50%. So, when we substitute the values, we obtain:

(267x-72/72)x100=50

5. Let's clear the "x":

(26700x-7200/72)=50
26700x-7200=50x72
26700x-7200=3600
26700x=3600+7200
x=10800/26700
x=$0.40
x=40 cents

What should the selling price per pound be?

The answer is: 40 cents or $0.40

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