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Paha777 [63]
3 years ago
12

Find the Area, I'm pretty sure you suppose to add them all up but my teacher is saying it's wrong

Mathematics
1 answer:
Alexxx [7]3 years ago
5 0
The answer to your question is 6.
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Hey need help on these NEED HELP ASAP
GalinKa [24]

Answer:

first one : (7+12+3+4+4+8)*2=76

8 0
3 years ago
Why should you record a one in the tens column when regrouping in addition?
ICE Princess25 [194]
To keep the place holder
5 0
3 years ago
5 (i + 2) = 8 (i -1)
Vinil7 [7]
5(i + 2) = 8(i -1)
5i + 10 = 8i -8
-3i + 10 = -8
-3i = -18
i = 6
5 0
3 years ago
What is 6 x 8 1/2<br><br> Please show me how to do it
Andrei [34K]
Answer:
= 6/1 × 17/2

= (6 × 17) / (1 × 2)

= 102/2

= 51/1

<span>= 51</span>
3 0
3 years ago
A sine function is transformed such that it has a single x-intercept in the interval (0,pi), a period of pi and a y-intercept of
Alex787 [66]
Just to make sure we're using the same language, I'm going to use the function form of:

y = A\sin(kx) + h

[] I would agree that k = 2, since the period is only half as long as a normal sine function. So, we so far, have y = A sin(2x) + h. We still need to find A and h.

[] The y-intercept is 3. Remember that the y-intercept happens when x = 0. So, plugging in x = 0 into our formula, we have: 3 = A sin(2*0) + h. In other words, 3 = A sin(0) + h = 0 + h = h. So, we now know that h = 3. The formula is now y = A sin(2x) + 3.

[] Finally, there is a single x-intercept. Picture what the sine function looks like right now, it is floating in the air around y = 3. We need to stretch it vertically until it just grazes the x-axis. 

If A = 1, then our sine function bounces between 2 and 4 (+/- 1 around h = 3). But that doesn't touch 0, so no good.

If A = 2, then our sine function bounces between 1 and 5 (+/- 2 around h = 3). Again, not quite touching 0 yet.

The answer should be A = 3, then our sine function bounces between 0 and 6 (+/- 3 around h = 3).

The final formula is y = 3 sin(2x) + 3.
7 0
3 years ago
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