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jasenka [17]
3 years ago
12

(x2−1)(x+1)≥0. pls help me​

Mathematics
1 answer:
docker41 [41]3 years ago
7 0

Step-by-step explanation:

(x²-1)(x+1)≥0

As it has two multiples,

Therefore,

Either,

x²-1≥0

x²≥0+1

x²≥1

x≤√1

x≤1

Or,

x+1≥0

x≥-1

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What is the mean set of the set of numbers
Arturiano [62]

Answer:

2

Step-by-step explanation:

-10 - 5 + 2 + 6 + 17 = 10

10 / 5 = 2

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3 years ago
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In the isosceles trapezoid below,<br> X =<br> 5x+ 15°<br> 7x-11°
tangare [24]

Answer:

x = 13°

Step-by-step explanation:

5x+15°=7x-11°

5x-7x = -11°-15°

-2x = -26° divide both sides by -2

x = 13°

5x+15° = 5×13°+15° = 80°

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4 0
3 years ago
*HELP*
vladimir1956 [14]

Answer:

The answer is 1/8

Step-by-step explanation:

Hope this helps. So the question is asking you odds of 8 which is the denominator already. So then it asks the odd to pick eight. So thats how i got 1/8.

5 0
3 years ago
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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c
Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 \rm{in^3}

Step-by-step explanation:

The linearization or linear approximation of a function f(x) is given by:

f(x_0+dx) \approx f(x_0) + df(x)|_{x_0} where df is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh

Taking R_0=1.5 inches and h=3 inches and evaluating the partial derivatives we obtain:

V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh

substituting the values and taking dx=0.1 and dh=0.3 inches we have:

V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}

Therefore the change in volume is estimated to be 17.20 \rm{in^3}

4 0
3 years ago
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