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NISA [10]
3 years ago
12

Two step inequalities khan Academy. Anybody know this answer?????

Mathematics
2 answers:
dimaraw [331]3 years ago
4 0

Answer:

C

Step-by-step explanation:

BartSMP [9]3 years ago
3 0
The answer to this is C.
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Hey yall i need help again im just lazy but will mark brainlyest
ElenaW [278]

Answer:  1=1 question 2 = 1 again 5-6 is the next and 4 3-10  5-6 next answer2-7  3-8 5-4 -22--9 2-29 are all the answers

Step-by-step explanation:

3 0
3 years ago
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Identify the postulate that proves the triangles are congruent.
Ratling [72]

Answer:

ASA

Step-by-step explanation:

Triangle ABD is congruent to triangle BDC by ASA because angle B in ABD  is 30 degrees. Then they give you BD which equals BD by the reflexive property. And finally angle D in ABD is also 30 degrees. In triangle BCD, angle B is 30 degrees, BD = BD, and angle D in BCD is 30 degress. So we have an angle (B), side (BD), and another angle (D)

6 0
3 years ago
Find the median of 8, 9, 9, 11, 12, 16.A) 8 B) 9 C) 10 D) 11
never [62]

Answer:

10

Step-by-step explanation:

11+9=20

20/2

7 0
3 years ago
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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.



3 0
3 years ago
Is the triangle isosceles, equilateral, or
lesantik [10]

Answer:

<em>B</em><em>=</em><em> </em><em>Equilateral</em>

<em>hope</em><em> </em><em>this helps</em><em>:</em><em>)</em>

4 0
2 years ago
Read 2 more answers
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