MATHEMATICS HIGH SCHOOL

Reese read twice as many pages Saturday than she read Sunday. If she read a total of 78 pages over the weekend, how many pages did Reese read Sunday?

Answers

Answer 1

Answer:

Given:

Reese read twice as many pages Saturday than she read Sunday.

She read a total of 78 pages over the weekend.

To find:

The number of pages she read on Sunday.

Solution:

Let x be the number of pages she read on Sunday.

Reese read twice as many pages Saturday than she read Sunday. So, the number of pager she read on Saturday is 2x.

Total number of page she read over the weekend is:

$Total=x+2x$

$Total=3x$

She read a total of 78 pages over the weekend.

$3x=78$

$x=(78)/(3)$

$x=26$

Therefore, Reese read 26 pages on Sunday.

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Tracy wants to earn at least 100 dollars from her two jobs next week. At most she can work 12 hours. Her first job pays 8 an hour and her second job pays 9 an hour. Let x represent the number of hours worked at the first job and y represent the number of hours worked at the second job. Which system of linear inequalities models tracy's equation

Answers

Answer:

Step-by-step explanation:

You didn't list the options from which we are to choose as your system of inequalities, but that doesn't matter...we'll come up with them on our own and then you can match them to your options. The first inequality is going to be about the number of hours worked. The second inequality is going to be about the money earned. Hours worked and money earned have to be in 2 different inequalities because they are not the same. If x is one job and y is the other, and the combination of these jobs cannot be more than 12 hours total, then the inequality for this is:

x + y ≤ 12

That represents the hours worked. As far as the money goes, she makes $8 per hour, x, at the first job, and $9 per hour, y, at the second job. She wants the combination of these wages to be at least $100. The inequality that represents the money earned is:

8x + 9y ≥ 100

That is the system that represents your situation.

Cory has 15 die-cast cars in his collection. Each year his collection increases by 20%. Roger has 40 cars in his collection. Each year he collects 1 additional car.Part A: Write functions to represent Cory and Roger's collections throughout the years.
Part B: How many cars does Cory have after 6 years? How many does Roger have after the same number of years?
Part C: After approximately how many years is the number of cars that Cory and Roger have the same? Justify your answer mathematically.

Answers

Part A:
For Roger: $y=x+40$
For Cory: $y=15 (1.2)^(x)$
where y is the total number of cars each boy collects over the year, x as the number of years passed.

Part B:
For Roger: 46
For Cory: 44.79~ 45 cars

Part C:
Here, we are trying to find the number of years (x) where Cory and Roger have the same number of cars.
By equating Roger and Cory's functions, we can solve for x
$x+40=15 (1.2)^(x)$ .
Since we cannot solve the value of x directly, we use trial and error to estimate the year.
When x=1
$41 \neq 18 \n$
When x=2
$42 \neq 21.6$
When x=3
$43 \neq 25.92$
When x=4
$44 \neq 31.10$
When x=5
45=/=37.3
When x=6
46=/=44.8
When x=7
47=/=53.7

The years that pass by before Cory and Roger have nearly the same number of cars is 6.

Answer:

Step-by-step explanation:

Part A:

For Roger:

For Cory:

where y is the total number of cars each boy collects over the year, x as the number of years passed.

Part B:

For Roger: 46

For Cory: 44.79~ 45 cars

In any triangle, there can be at most 1 obtuse angle. At most, how many obtuse angles can there be in an equilateral triangle?

Answers

Zero. An equilateral triangle has only 60 degree angles

Can someone help me with this and explain me how to do it please :) ^

Answers

Answer:

Step-by-step explanation:

2√x, √x + 1, 2√x + 1 sides of triangle

√x= y

2y, y+1, 2y+1

(2y)²+(y+1)²= (2y+1)²

4y²+y²+2y+1= 4y²+4y+1

y²-2y=0

y=0 (discarded as side can't be zero) or y= 2

so √x=2 ⇒ x= 4

Answer:

x= 4

$perpendicular (p) = √(x) + 1$

$base(b) = 2 √(x)$

$hypotenuse(h) = 2 √(x) + 1$

According to the Pythagoras Theoram,

p²+b²= h²

${( √(x ) + 1) }^(2) + {(2 √(x) )}^(2) = {(2 √(x) + 1)}^(2)$

=>x+2√x+1+4x = 4x +4√x +1

=> x = 4x - 4x +4√x- 2√x+ 1- 1

=>x= 2√x

=> x/√x = 2

=> √x = 2

Therefore, x= (√x)² = (2)² = 4

What is the answers to this math 3-3x6+2=

Answers

Follow the BDMAS rule: -3x6= -18. 3-18+2. 3-16 = -13 is the answer. :))

3-3x6+2= 3 times 6 is 18 3-18= -15 -15+2=-13 So I used PEMDAS with this problem I hope this help if u want more information tell me so I can help

Liquid A has a density of 0.7g/cm^3liquid B has a density of 1.6g/cm^3

140g of liquid A is and 128g of liquid B are mixed to make liquid C

work out the density of liquid C

Answers

density is defined as mass per unit volume

in this case liquids A and B with 2 different densities are mixed and we are asked to find the density of the liquid C

liquid A has a density of 0.7 g/cm³

mass of liquid A added is 140 g

therefore volume of liquid A added is - 140 g / 0.7 g/cm³ = 200 cm³

liquid B has a density of 1.6 g/cm³

mass of liquid B added is 128 g

volume of liquid B added is - 128 g / 1.6 g/cm³ = 80 cm³

the total mass of liquid C after adding liquid A and B = 140 g + 128 g = 268 g

total volume in liquid C - 200 cm³ + 80 cm³ = 280 cm³

density of liquid C = mass / volume

= 268 g / 280 cm³ = 0.957 g/cm³

density of liquid C - 0.957 g/cm³

$\rho = (m)/(V) \Rightarrow V=(m)/(\rho)$
ρ - density, m - mass, V - volume

Liquid A:
$\rho=0.7 \ (g)/(cm^3) \nm=140 \ g \n\n V=(140 \ g)/(0.7 \ (g)/(cm^3))=200 \ cm^3$

Liquid B:
$\rho=1.6 \ (g)/(cm^3) \nm=128 \ g \n\n V=(128 \ g)/(1.6 \ (g)/(cm^3))=80 \ cm^3$

Liquid C:
$m=140 \ g + 128 \ g=268 \ g \nV=200 \ cm^3 + 80 \ cm^3=280 \ cm^3 \n \n\rho=(268 \ g)/(280 \ cm^3) \approx 0.96 \ (g)/(cm^3)$

The density of liquid C is approximately 0.96 g/cm³.

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