Ask question

Ask question

All categories

2 years ago

5

Eric has challenged himself to walk 24,000 steps in 4 days. If Eric walks the same number of steps each day, which function repr

esents the number of steps Eric still needs to walk to reach his goal with respect to the number of days since he started his challenge?
y = 8,000x − 24,000
y = -8,000x + 24,000
y = 6,000x − 24,000
y = -6,000x + 24,000

1 answer:

Alborosie2 years ago

4 0

If Eric wanted to walk 24,000 steps in 4 days, he would have to walk 6,000 steps each day (24/4=6). The correct answer would be the bottom answer as saying he has already walked 2 days it would be (y=-6,000(2)+24,000) this would simply out to (y=-12,000+24,000) and then (y=12,000).

You might be interested in

Emma earns a $39,000 salary in the first year of her career. Each year, she gets a 5% raise. How much does Emma earn in total in

jeyben [28]

Answer: 21

Step-by-step explanation: like basically

4 0

2 years ago

Read 2 more answers

The table shows ordered pairs of the function y = 16 + 0.5x . Which ordered pair could be the missing values represented by (x,

lawyer [7]

The <u><em>correct answer</em></u> is:

(8, 20).

Explanation:

We can substitute the values for x in the equation and see if they equal y:

y=16+0.5(0) = 16+0 = 16 ≠ 18. This is not the correct option.

y=16+0.5(5) = 16+2.5 = 18.5 ≠ 19.5. This is not the correct option.

y=16+0.5(8) = 16+4 = 20. This is the correct option.

3 0

2 years ago

Read 2 more answers

For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How

natulia [17]

<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let $e$ the cost of the exam hours

Let $w$ be the cost of the workshop hours

Let $l$ be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

$e=2w$ equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

$w=2l$ equation (2)

Finally, we also know that 3hr exams 24hr workshops and 12hr lectures cost $528, so:

$3e+24w+12l=528$ equation (1)

Now, lets find the value of $l$ :

Step 1. Solve for $l$ in equation (3)

$3e+24w+12l=528$

$12l=528-3e-24w$ equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

$12l=528-3e-24w$

$12l=528-3(2w)-24w$

$12l=528-6w-24w$

$12l=528-30w$ equation (5)

Step 3. Replace equation (2) in equation (5) and solve for $l$

$12l=528-30w$

$12l=528-30(2l)$

$12l=528-60l$

$72l=528$

$l=\frac{528}{72}$

$l=\frac{22}{3}$

$l=7.33$

Cost of lectures = $7.33 per hour

3 0

2 years ago

Read 2 more answers

I need to write a decimal where the value of a digit is 1/10 as much as a digit in another place

KengaRu [80]

1/10 is the same as saying 10 times less

Refer to the diagram below
The digit as the first decimal place is worth 1/10
The digit as the second decimal place is worth 1/100 which is 1/10 worth the digit as the first decimal place

Example of two decimal numbers:

2.56 and 2.68

The digit 6 in 2.56 is 6/100
The digit 6 in 2.68 is 6/10

The digit 6 in 2.56 is 1/10 as much as the digit 6 in 2.68

6 0

2 years ago

Sum of positive roots of the equation (x2 - 12x + 35) (x2 + 10x + 24) = 504 is

Digiron [165]

Answer:

(3)11

Step-by-step explanation:

We are given that

$(x^2-12x+35)(x^2+10x+24)=504$

We have to find the sum of positive roots of the equation.

$x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504$

$x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504$

$x^4-2x^3-61x^2+62x+840-504=0$

$x^4-2x^3-61x^2+62x+336=0$

Factor of 336

2,3,4,6,8,7,

Let x=2

$2^4-2^4-61(2^2)+62(2)+336\neq0$

x=2 is not the root of equation

x=-2

$(-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0$

Hence x=-2 is the root of equation.

x+2 is a factor of equation.

x=3

$3^4-2(3^3)-61(3^2)+62(3)+336=0$

Therefore, x=3 is the root of equation.

$(x+2)(x-3)(x^2-x-56)=0$

$(x+2)(x-3)(x^2-8x+7x-56)=0$

$(x+2)(x-3)(x(x-8)+7(x-8))=0$

$(x+2)(x-3)(x-8)(x+7)=0$

$x-8=0\implies x=8$

$x+7=0\implies x=-7$

Positive roots are 3 and 8

Sum of positive roots=3+8=11

Option (3) is true.

4 0

2 years ago