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taurus [48]
2 years ago
6

A:40 B:70 C:80 D:120 E:150

Mathematics
1 answer:
lara31 [8.8K]2 years ago
7 0

Answer:

E: 150

Step-by-step explanation:

Remember, a straight line is 180 degrees, and the total degrees in a triangle is 180. Also, the angle is clearly an obtuse angle, so just get rid of A, B, C, lol.

I think it would be more effective to show you this picture I made, but if you have any questions, I’ll try to answer them.

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Analyze the data represented in the table and select the appropriate model
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Answer: Option B, a linear relation.

Step-by-step explanation:

Let's analyze the points in the data:

the first 3 pairs are: (0,14.70) (10,19.03) and (20, 23.36)

We can see the differences between consecutive points and see if the relation is linear:

(10,19.03) -  (0,14.70) = ( 10 - 0, 19.03 - 14.70) = (10, 4.33) now we need to see the quotient: 4.33/10 = 0.433

(20, 23.36) - (10,19.03)  = (20 - 10, 23.36 - 19.03) = (10, 4.33), and the quotient is the same as before.

This means that the relation is a linear relation, where the slope os 0.433 and the x-intercept is 14.70, so the equation can be written as:

y(x) = 0.433*x + 14.70

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Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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son4ous [18]
I just took the test yesterday and passed
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