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2 years ago

5

Justin has 180 toothpicks. Katie says, "Justin has four times as many toothpicks as I do!" Which equation could represent this s

ituation?

1 answer:

ICE Princess25 [194]2 years ago

4 0

Multiplication would represent the equation because in order to find out how many toothpicks just has you would need to MULTIPLY to find out.

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Round 69,814 to the nearest hundred

tamaranim1 [39]

Answer:

69,800 thats the answer

6 0

2 years ago

One package of tile will cover 3 square feet.How many packages will shee need 8,13,15,39

IrinaVladis [17]

Hey there! :)

Answer:

13 packages.

Step-by-step explanation:

Begin by finding the total area of this composite figure. Separate the figure into separate rectangles. Use the formula A = l × w to calculate the area of each:

Smaller rectangle:

Subtract 7 from 9 to find the width:

9 -7 = 2. Therefore:

2 × 2 = 4 ft².

Larger rectangle:

7 × 5 = 35 ft².

Add up the two areas to find the area of the entire figure:

4 + 35 = 39 ft².

If each package of tile covers 3 ft², simply divide to find the package of tiles needed:

39 / 3 = 13 packages.

4 0

3 years ago

(Very easy) Find the volume. Round to the nearest tenth if necessary.

Pavel [41]

Answer:

410.5 yards cubed

Step-by-step explanation:

Volume of a cone is 1/3Bh

h is the height of the cone

B is the area of the base, which is a circle, so use πr^2 to find area of the circle

The radius is 7, so:

π7^2

π49(8)(1/3) = 410.5

6 0

2 years ago

Read 2 more answers

If a1=10 and an=an-1+4 then find the value of a5

tamaranim1 [39]

Answer:

26

Step-by-step explanation:

Each term is the previous with 4 added to it.

a1 = 10, a2 = 14, a3 = 18, a4 = 22, a5 = 26

3 0

2 years ago

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t

WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

$\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}$

Seperate the differential equation and solve for the constant C.

$\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}$

You have 100 rodents when:

$100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months$

You have 1000 rodents when:

$1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months$

7 0

2 years ago