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Pani-rosa [81]
3 years ago
12

Identify the radius of the circle whose equation is (x - 2)^2 + (y - 8)^2 = 16.

Mathematics
2 answers:
KiRa [710]3 years ago
5 0
The radius is 4
the equation of a circle is (x-h)^2+(y-k)^2=r^2
in this case, 16=r^2
so r=4
Misha Larkins [42]3 years ago
3 0

16^2 is the radius, so 16 squared is 4. Therefore the radius is equal to 4

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Bianca filled 2 same-sized jars with flavored popcorn as a gift. She wants to glue a piece of ribbon around the edge of each lid
dybincka [34]
I think it is 177.4728 

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7 0
3 years ago
Read 2 more answers
What does x equals in 2x + 16= -26
vredina [299]

Answer:

-21

Step-by-step explanation:

-21 x 2 =  -42                   -42+16= -26

7 0
3 years ago
Let c = 10.Which expression makes the comparison true.
den301095 [7]

Answer:

c² - 63

Step-by-step explanation:

4c - 10 = 30

3c ÷ 3 = 10

c² - 63 = 37

c² ÷ 10 = 10

Since c² - 63 = 37 and c + 25 = 35, 37 is greater than 35 so it would make sense to choose c² - 63 as the answer.

c + 25 < 37

I hope this helps you :D

3 0
3 years ago
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
If p(x) = x^2 + 7x + 3 is divided by x + 4, the remainder is <br> a0
Sphinxa [80]
Hope this helps you!

4 0
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