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aniked [119]
3 years ago
12

Plz help me solve this ASAP and show the work thank you

Mathematics
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

14

Step-by-step explanation:

The diagram is right angle triangle so we can use SOHCAHTOA

So in the diagram we have 62 degree opposite to x and the hypotenuse

Step 1

Sin62=x/16

Step 2

X=16sin62 by cross multiplication

Step 3

X=14.13

X=14

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What does -22 belong to
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Answer:

it's part of the real number system

it's a rational number

it's an integer

Step-by-step explanation:

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2 years ago
The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per
posledela
136,000,000/31,000= how many hours for the spaceship to reach mars
136,000,000/31,000/24= how many days for the spaceship to reach mars
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7 0
3 years ago
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which is a shrink of an exponential growth function? a. f(x) = 1/3(3)^x b. f(x) = 3(3)^x c. f(x) = 1/3(1/3)^x d. f(x) = 3(1/3)^x
Gnesinka [82]

Answer: option a.

f(x)=\frac{1}{3} (3)^x


Explanation:


A <em>shrink</em> of a function is a <em>shrink</em> on the vertical direction. It means that for a certain value of x, the new function will have a lower value, in the intervals where the function is positive, or a higher value, in those intervals where the function is negative. This is, the image of the new function is shortened in the vertical direction.


That is the reason behind the rule:

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So, we just must apply the rule: to find a shrink of an exponential growth function, multiply the original function by a scale factor less than 1.


Since it <em>is a shrink of</em> <em>an exponential growth function</em>, the base must be greater than 1. Among the options, the functions that meet that conditon are a and b:


a. f(x)=\frac{1}{3} (3)^x \\ \\ b.f(x) = 3(3)^x


Now, following the rule it is the function with the fraction (1/3) in front of the exponential part which represents a <em>shrink of an exponential function</em>.

8 0
3 years ago
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sashaice [31]

Answer:

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x = 2

4 0
1 year ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
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To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

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Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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