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Korvikt [17]
2 years ago
15

What is the greatest whole number that rounds to 2,800 when rounded to the nearest​ hundred? The least whole​ number?

Mathematics
1 answer:
zhannawk [14.2K]2 years ago
3 0

Answer:

The greatest is 2,799

The lowest is 2,750

Step-by-step explanation:

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4 0
2 years ago
Which graph represents f(x)=6cos(4πx) ?
lesya692 [45]

Answer:

The graph representing f(x)=6cos(4πx) is shown in the attached file.

Step-by-step explanation:

The graph representing f(x)=6cos(4πx) is shown in the attached file.

6 0
3 years ago
What is 3 equivalent ratios for 2/6
Crazy boy [7]
1/3
3/9
4/12
are equivalent to 2/6
3 0
3 years ago
Find the circumference of the circle shown. Round to the nearest tenth.
Sunny_sXe [5.5K]

Answer:

110 cm

Step-by-step explanation:

So if we know the diameter, which is 35 cm, we can use the formula π*d. So we would take the diameter and put it in the place of d. So it would be pie times 35. Which would result in us getting 109.96. And this rounded to the nearest tenth would be 110. So The circumference would be 110 cm.

5 0
3 years ago
Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?
Cloud [144]

Answer:

True; quadrants I & IV

Step-by-step explanation:

We know the relation between sine and cosine function which is given by

\sin^2 \theta +\cos^2 \theta = 1

Let us solve this equation for cosine function.

\cos^2 \theta = 1-\sin^2 \theta

Take square root both sides. When ever we take square root we need to write the solution in plus minus form

\sqrt{\cos^2 \theta}=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=-\sqrt{1-\sin^2 \theta}, \sqrt{1-\sin^2 \theta}

If Θ is in quadrants I and IV then the value will be positive and if Θ is in II and III quadrant then the value is negative.

Hence, if Θ is in quadrants I & IV, then we have

\cos \theta=\sqrt{1-\sin^2 \theta}

Thus, the correct option is: True; quadrants I & IV


6 0
3 years ago
Read 2 more answers
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