Math Word Problems Homework Help!?

1. 1st number—x:

x + x + 1 + x + 2 = 297

3x = 294

x = 98

Answer: 98, 99, 100 are the numbers.

2. Tens digit—x:

10x + x - 4 < 8(x + x - 4)

11x - 4 < 8(2x - 4)

11x - 4 < 16x - 32

28 < 5x

x > 28/5 or 5.6

The number:

= 10(6) + 6 - 4

= 60 + 2

= 62

Answer: 62 must be one of the numbers.

3. Age of the daughter—x:

x + 18 + 10 + 2 = 2(x + 10)

x + 30 = 2x + 20

x = 10

Age of the man:

= 10 + 18

= 28

Answer: The man is 28, his daughter is 10.

4. Rate of the slower car—x:

1.25(x + x + 10) = 212.5

2x + 10 = 170

x + 5 = 85

x = 80

Rate of the faster car:

= 80 + 10

= 90

Answer: 90 kph & 80 kph are their rates.

5. Maximum amount on food—x:

x + 2x = 0.6(Php32,000.00)

3x = Php19,200.00

x = Php6,400.00

Answer: Php6,400.00 monthly food expense.

6. Width—x:

x(x + 2) = 24

x² + 2x = 24

x² + x = 24 + 1²

x² + x = 24 + 1

(x + 1)² = 25

x + 1 = 5

x = 4

Length:

= 4 + 2

= 6

Answer: 4 inches by 6 inches are the dimensions.

1. Let the three numbers be x, x + 1 and x + 2. their sum is equal to 297, hence

x + (x + 1) + (x + 2) = 297

Solve for x and find the three numbers

3x+3 =297

3(x+1) =297

X+1 = 297/3 = 99

x= 99-1 =98

so numbers are

x= 98

x+1 = 99

x=2 = 100

ans ===== 98,99,100

3. Let daughter be x yrs old. Father is 18 years older than daughter. In 10 years, father will

be two times as old as daughter and subtract 2 years. How old is daughter now?

Let x be daughter’s age now. Add 10 to get the ages in 10 yrs.

age now age in 10 yrs

daughter x x+10

Father x+18 x+18+10

Write the new relationship in an equation using the ages in 10 yrs.

In 10 years, father will be twice as old as daughter and less 2 yrs.

x + 28 = 2(x + 10)-2

x + 28 = (2x + 20) - 2

x + 28 - 2x - 20 = -2

-x +8 = -2

-x = -2-8

-x= -10 ... x = 10

daughter's age x = 10

fathers age x+18+10 ==== 10+18+10 = 38

2. let x be units digit

let y be tens didgit

The unit's digit of a two-digit number is 4 less than it's ten's digit

so x = y - 4

then yx ----->

number is less than 8 times the sum of it's digit.

( y + y - 4) - 8 = 0

2y - 4 - 8 =0

2y = 12

y = 6

x = y-4 = 6-4 =2

ans the number is 62

1. The three numbers, we'll call them a, b, c, need to add to 297, so: a+b+c=297. But they must be consecutive, so you know that b is 1 greater than a and c is 2 greater than a. So...b=a+1 and c=a+2. Go from there and solve a+b+c=297 for a. (3a+3=297)

2. You only have so many options, i believe: 95, 84, 73, 62, 51, 40.

3. Use algebra, like in question 1. Say m=man's age and d=daughter's age. Make 2 equations, the first being m=d+18, the second is (i think?), (m+10)=2(d+10)-2. You have 2 equations and 2 unknowns, now substitute and solve.

I gotta get to sleep...hope that helps with the first couple.

1. 98, 99, 100 98 +99 is 197 + 100 is 297.

2. 84 because 8+4 is 12 x 8 is 96 and 84 is less than 96

3. he is 48 and she is 30 because 18 more than 30 is 48. in ten years he'll be 58, double her age is 60.

4. not sure

5. 12,800 on food and 6,400 on education. 60% of 32,000 is 19,200. 6,400+ 12,800 is 19,200.

6. width 4 length 6

why r u getting hw its august?

say (2/5 ) x ( 3 and 4/5 ) 1st simplify mixed fraction => ( 2/5) x ( 19/5) then multiply => 38 / 25 Clear ?

1. 98+99+100

6. length=6 width=4

Trev-O...learn to add.