Select the correct answer.

Nadusha1986 [10]

ANSWER:
12x
“A number” just means a variable normally

5 0

3 years ago

Reggie and jay go for a walk morning.reggie walks 2 1/4 miles. Jay walks 1 3/8 miles less than Reggie. What is the total distanc

jek_recluse [69]

Jay walks 2 1/4 miles which if you double it so the denominators are e same for both 2 1/4 and 1 1/8 then your answer will be 18/8 (Jay) and 9/8 (Reggie). Meaning that Reggie walks have the distance Jay walks.

7 0

4 years ago

Can someone help me please this is my last one

neonofarm [45]

Answer:

Domain and range: [0, ∞)

Step-by-step explanation:

There is no limit to the number of concert tickets someone can buy, but there cannot be negative concert tickets. The domain, or what the input can be, is [0, ∞) as it can be greater than or equal to 0

The range is the total cost, or what the output can be. Since the total cost is the number of tickets multiplied by 55 (as it is $55 per ticket), this can be represented by 55 * number of tickets. The total cost cannot be negative, but it can be 0 if no one buys tickets. As there is no limit to the amount of tickets someone can buy, there is no limit to what the total cost could be, making the range [0, ∞)

3 0

3 years ago

I left Boston for Washington, DC, at 7:00 am. At first, my speed was 55 mph; then it was 75 mph. I traveled 630 miles in 10 hour

borishaifa [10]

Answer:

We began to drive at 75 mph at 1:00 p.m.

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Speed at the beginning of the trip = 55 mph

Speed at the end of the trip = 75 mph

Distance from Boston to Washington D.C. = 630 miles

Time of the trip = 10 hours

2. At what time did I begin to drive faster?

Let's recall that Speed = Distance/Time, therefore Time = Distance/Speed

Let x represent the distance we drove at 55 mph and 630 - x the distance we drove at 75 mph, thus:

Time we drove at 55 mph + Time we drove at 75 mph = Time of the trip

Replacing with the real values, we have:

x/55 + (630 - x)/75 = 10

75x + 55 * (630 - x) = 41,250 (4,125 as Lowest common multiple)

75x + 34,650 - 55x = 41,250

20x = 41,250 - 34,650 (Subtracting 34,650 at both sides)

20x = 6,600

x = 6,600/20

x = 330 ⇒ 630 - x = 300

Time we drove at 55 mph = 330/55 = 6 hours

Time we drove at 75 mph = 300/75 = 4 hours

We began to drive at 75 mph at 1:00 p.m. (7 am + 6 hours)

7 0

3 years ago

HEY‼️‼️ CAN SOMEONE HELP WITH THIS⁉️ 1 and 4 are completed I need 2 3 5 6 7 8 9 USE THE FORMULA SA= ph + 2b B= base p= perimeter

Oksana_A [137]

Answer:

Part 1) $SA=58\ cm^2$

Part 2) $SA=90\ ft^2$

Part 3) $SA=168\ m^2$

Part 4) $SA=96\ in^2$

Part 5) $SA=660\ in^2$

Part 6) $SA=96\ cm^2$

Part 7) Your friend is not correct, $SA=150\ cm^2$

Part 8) $SA=486\ in^3$

Part 9) $SA=94\ ft^2$

Step-by-step explanation:

we know that

The surface area of a rectangular prism is equal to

$SA=2B+Ph$

where

B is the area of the base of the prism

P is the perimeter of the base of the prism

h is the height of the prism

Part 1) Find the surface area of the rectangular prism

Find the area of the base B

$B=(4)(1)=4\ cm^2$

Find the perimeter of the base

$P=2(4+1)=10\ cm$

we have

$h=5\ cm$

substitute the values in the formula of surface area

$SA=2(4)+10(5)=58\ cm^2$

Part 2) Find the surface area of the rectangular prism

Find the area of the base B

$B=(2)(5)=10\ ft^2$

Find the perimeter of the base

$P=2(2+5)=14\ ft$

we have

$h=5\ ft$

substitute the values in the formula of surface area

$SA=2(10)+14(5)=90\ ft^2$

Part 3) Find the surface area of the rectangular prism

Find the area of the base B

$B=(9)(6)=54\ m^2$

Find the perimeter of the base

$P=2(9+6)=30\ m$

we have

$h=2\ m$

substitute the values in the formula of surface area

$SA=2(54)+30(2)=168\ m^2$

Part 4) Find the surface area of the triangular prism

Find the area of the triangular base B

$B=(1/2)(8)(3)=12\ in^2$ ---> the area of triangle

Find the perimeter of the triangular base

$P=8+5+5=18\ in$

we have

$h=4\ in$

substitute the values in the formula of surface area

$SA=2(12)+18(4)=96\ in^2$

Part 5) Find the surface area of the triangular prism

Find the area of the triangular base B

$B=(1/2)(12)(5)=30\ in^2$ ---> the area of triangle

Find the perimeter of the triangular base

$P=12+5+13=30\ in$

we have

$h=20\ in$

substitute the values in the formula of surface area

$SA=2(30)+30(20)=660\ in^2$

Part 6) Find the surface area of the triangular prism

Find the area of the triangular base B

$B=(1/2)(6)(8)=24\ cm^2$ ---> the area of triangle

Find the perimeter of the triangular base

$P=8+6+10=24\ cm$

we have

$h=2\ cm$

substitute the values in the formula of surface area

$SA=2(24)+24(2)=96\ cm^2$

Part 7) we know that

The surface area of a cube is equal to the area of its six square faces

$SA=6b^2$

where

b is the length side of the cube

we have

$b=5\ cm$

substitute

$SA=6(5^2)$

$SA=150\ cm^2$

therefore

Your friend is not correct

The surface area of the cube is $SA=150\ cm^2$

Part 8) we know that

The surface area of a cube is equal to the area of its six square faces

$SA=6b^2$

where

b is the length side of the cube

we have

$b=9\ in$

substitute

$SA=6(9^2)$

$SA=486\ in^3$

Part 9) Find the surface area of the rectangular prism (shipping crate)

Find the area of the base B

$B=(5)(4)=20\ ft^2$

Find the perimeter of the base

$P=2(5+4)=18\ ft$

we have

$h=3\ ft$

substitute the values in the formula of surface area

$SA=2(20)+18(3)=94\ ft^2$

6 0

3 years ago