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sesenic [268]
3 years ago
11

Erika says that no matter how many decimal places she divides to when she divides 1 by 3, the digit 3 in the quotient will just

keep repeating. Is she correct? Explain
Mathematics
1 answer:
inn [45]3 years ago
8 0
Yes she is , because when you divide 1 into 3 the answer is 0.3 and it’s repeating !
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Daniel buys two apples for every five oranges. If you bought 12 apples, how many oranges dis he buy?
wariber [46]

Answer:

30 oranges

Step-by-step explanation:

2 apples : 5 oranges

12 apples : 30 oranges

because 12 apples is 6 times the original ratio, so we multiply the 5 oranges by 6 too.

8 0
3 years ago
Determine if side lengths 24, 21.5, and 55.5 make a triangle. If so, classify the type of triangle it creates.
valentina_108 [34]

Answer

b

Step-by-step explanation:

8 0
3 years ago
I’m really struggling, someone please help!
Ulleksa [173]

Hi there! :)

Answer:

\huge\boxed{C.}

We can examine each answer choice individually:

A. 569 × 10² = 569 × (10 · 10) = 569 · 100 = 56,900. Therefore, this choice is incorrect.

B. 569 · 10 = 5,690. This choice is incorrect.

C. 10³ · 569 = (10 · 10 · 10) ·569 = 1000 · 569 = 569,000. This choice is correct.

D. 10² · 569 = (10 · 10) · 569 = 56,900. This choice is incorrect.

Therefore, the correct option is C.

6 0
3 years ago
Read 2 more answers
Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.
WARRIOR [948]

Answer:

3rd option

Step-by-step explanation:

Using the identities

cot x = \frac{1}{tanx}

csc² x = 1 + cot² x

Given

tanθ = \sqrt{3} , then cotθ = \frac{1}{\sqrt{3} }

csc²θ = 1 + (\frac{1}{\sqrt{3} } )² = 1 + \frac{1}{3} = \frac{4}{3}

cscθ = ± \sqrt{\frac{4}{3} } = ± \frac{2}{\sqrt{3} }

Since θ is in 3rd quadrant, then cscθ < 0

cscθ = - \frac{2}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = - \frac{2\sqrt{3} }{3}

8 0
3 years ago
A point located at (4, 7) is translated up 5 units. What are the coordinates of the image?
Fantom [35]
Can u make a graaph showing this
7 0
4 years ago
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