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anzhelika [568]
3 years ago
5

What is 2+1 please help me

Mathematics
2 answers:
ANTONII [103]3 years ago
7 0

Answer:

ummmm i think its 3 bb

Step-by-step explanation: if i think really hard i think 2 + 1 is 3

erica [24]3 years ago
4 0
2+1=3

I hope this helped! :)
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How many coefficients are in the expression 5x³ - 2n² + 6x - 4?
ikadub [295]

Answer: 3 cofficients are there ( 5, 2, and 6).

8 0
3 years ago
If the period of a sine function is the number of units before the cycle begins to repeat, why would the period of the regressio
Jlenok [28]

Answer:

This means that after going around the unit circle once (2π radians), both functions repeat. So the period of both sine and cosine is 2π . Hence, we can find the whole number line wrapped around the unit circle.

Step-by-step explanation:

5 0
3 years ago
The function L = 0.8T^2 models the relationship between L, the length in feet of a pendulum, and T, the period in seconds of the
Semmy [17]
The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft

L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
7 0
3 years ago
In circle O, what is m∠MAJ? °
mr_godi [17]

See the attached figure.

m ∠KAJ = 170° &  m ∠LAM = 80°

We should know that :

m ∠KAJ + m ∠LAM + m ∠KAL + m ∠MAJ = 360°

∴ m ∠KAL + m ∠MAJ = 360° - (m ∠KAJ + m ∠LAM)

∴ m ∠KAL + m ∠MAJ = 360° - (170°+80°) = 360° - 250° = 110°

But : m ∠KAL = m ∠MAJ ⇒⇒⇒ <u>Opposite angles.</u>

∴ m ∠MAJ + m ∠MAJ = 110°

∴ 2 * m ∠MAJ = 110°

∴ m ∠MAJ = 110° ÷ 2 = 55°


<u>So, the answer is :  m ∠MAJ = 55°</u>

4 0
3 years ago
Assume the readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees
lozanna [386]

Answer: 0.0035

Step-by-step explanation:

Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.

i.e.  \mu=0 and \sigma= 1  

Let x denotes the readings on thermometers.

Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_

P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035

Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035

The required region is attached below .

8 0
3 years ago
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