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olga_2 [115]
3 years ago
7

Find the measure of each acute angle.

Mathematics
2 answers:
Rudik [331]3 years ago
7 0
X =5 so 11x-2=53 and 6x+7=37
Kruka [31]3 years ago
3 0

Answer:

11x-2 is 53 degrees and the other one is 37

Step-by-step explanation:

(11x-2)+(6x+7)=90 degrees 17x plus 5 is 90 and 90 minus is 85 and 85/17 is 5 and plug it back into the equations

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Answer all please, thank you :)
prisoha [69]

The angles in a triangle add up to 180 degrees.  That's the same sum as a linear pair, which we get when two lines cross.

1. 180 - 65 - 45 = 70

Answer: 70°

2. 180 - 90 - 35 = 55

Answer: 55°

3. 180 - 120 = 180 - 50 - x

x = 70

Answer: 70°

6 0
3 years ago
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My bed is 3 inches by 2 2/3 inches by 1/3 inches, what is the volue of my bed.
Romashka-Z-Leto [24]

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2.6

Step-by-step explanation:

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3 years ago
If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?
ivann1987 [24]

Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function f(x) = |(x^2-9)(x^2 + 1)|   in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}

Solution : f(x) is a function that satisfies all of the following :

1) f(x)  is continuous on the closed interval [-1,1]  

\lim_{x\to a} f(x)=f(a)

2) f(x) is differentiable on the open interval  (-1,1)

Then there is a number  c such that  f'(c)=\frac{f(b)-f(a)}{b-a}

f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16

f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16

f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0

f'(c)=0 ........[1]

Now, we find f'(x)

f(x) = |(x^2-9)(x^2 + 1)|

f(x) =x^4-8x^2-9

Differentiating w.r.t  x

f'(x) =4x^3-16x

In place of x we put x=c

f'(c) =4c^3-16c

f'(c) =4c^3-16c=0  (by [1], f'(c)=0)

4c(c^2-4)=0

4c=0,c^2-4=0

either c=0 or  c^2-4=0\rightarrow c=\pm2

we cannot take c=\pm2  because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.



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Katena32 [7]

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