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2 years ago

10

How many hundreds are there in 5000

2 answers:

dimulka [17.4K]2 years ago

8 0

We know that there are ten 100s in 1000 so there must be fifty 100s in 5000.
Hope this helps :)

Mila [183]2 years ago

6 0

The answer is 50
the easy way to find the answer is to take off the two zeros at the end of the number.

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Which ratio is equvalent to 7:3

Gala2k [10]

Answer:

21:18

Step-by-step explanation:

3 0

2 years ago

How many 1/4 cups goes into 2 2/4 cups

Serhud [2]

1/4 goes into 2 2 /4

Means 2 2/4 divided by 1/4

Change 2 2 /4 to improper fraction. = (2*2 + 4)/4 = 10/4

10/4 ÷ 1/4

10/4 × 4/1

= 10

So it is 10.

7 0

2 years ago

Read 2 more answers

The radius of a circle is 7 mm. What is the area.

Solnce55 [7]

Answer:

A= 153.94 mm

Step-by-step explanation:

A = πr²

A=π(7)²

A= 153.94 mm

7 0

1 year ago

Line segment AB has endpoints A(-1.5, 0) and B(4.5 8). Point C is on line segment AB located at (0, 2)

Pie

Answer:

The ratio of AC/CB=1/3

Step-by-step explanation:

We use the section formula:

$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

We substitute the point A(-1.5,0) and B(4.5,8).

We plug in the points to get:

$(\frac{4.5m - 1.5n}{m+n},\frac{8m+n \times 0}{m+n})$

This is supposed to be equal to: (0,2).

$(\frac{4.5m - 1.5n}{m+n},\frac{8m+n \times 0}{m+n}) = (0,2)$

We can equate, corresponding coordinates to find m and n.

$\frac{4.5m - 1.5n}{m + n} = 0$

This implies that:

$4.5m - 1.5n = 0$

$4.5m = 1.5n$

$\frac{m}{n} = \frac{1.5}{4.5}$

$\frac{m}{n} = \frac{1}{3}$

Therefore the ratio is m:n=1:3

4 0

2 years ago

7. Are the ordered pairs below an arithmetic sequence or geometric

DochEvi [55]

Answer:

8. Arithmetic Progression

9. $f(9) = 300$

Step-by-step explanation:

Given

$\{(1,55)\ (2, 45)\ (3, 35)\ (4, 25)\}$

Solving (8): Arithmetic or Geometric

We start by checking if it is arithmetic by checking for common difference (d).

$d = y_2 - y_1 = y_3 - y_2 = y_4 - y_3$

This gives:

$d = 45 - 55 = 35 - 45 = 25 - 35$

$d = -10 = -10 = -10$

$d=-10$

<em>Because the common difference is equal, then it is an arithmetic progression</em>

<em></em>

Solving (8):

$f(n) = f(1) + f(n-1)$

To find f(9), we substitute 9 for n

$f(9) = f(1) + f(9-1)$

$f(9) = f(1) + f(8)$

We need to solve for f(8); substitute 8 for n

$f(8) = f(1) + f(8 - 1)$

$f(8) = f(1) + f(7)$

We need to solve for f(7); substitute 7 for n

$f(7) = f(1) + f(7 - 1)$

$f(7) = f(1) + f(6)$

We need to solve for f(6); substitute 6 for n

$f(6) = f(1) + f(6 - 1)$

$f(6) = f(1) + f(5)$

We need to solve for f(5); substitute 6 for n

$f(5) = f(1) + f(5 - 1)$

$f(5) = f(1) + f(4)$

From the function, f(4) = 25 and f(1) = 55.

So:

$f(5)=55 + 25$

$f(5)=80$

$f(6) = f(1) + f(5)$

$f(6) = 55 + 80$

$f(6) = 135$

$f(7) = f(1) + f(6)$

$f(7) = 55 + 135$

$f(7) = 190$

$f(8) = f(1) + f(7)$

$f(8) = 55 + 190$

$f(8) = 245$

$f(9) = f(1) + f(8)$

$f(9) = 55 + 245$

$f(9) = 300$

8 0

2 years ago